How can I get common key from master secret? I have read RFC 5246 6.3. But I can't figure out;In TLS1.2 and TLS_RSA_WITH_AES_128_CBC_SHA;

Question

I am studying TLS1.2;

So,I launched a simple https server by Java that replies only "abc" and uses CipherSuites TLS_RSA_WITH_AES_128_CBC_SHA;

Certificate info is

RSA Private-Key: (2048 bit, 2 primes)
modulus:
    00:c1:0d:4c:ee:50:cf:8b:05:9c:e0:98:a6:11:b8:
    3b:49:34:63:07:c5:ce:ff:23:b2:2a:fa:e6:64:4f:
    34:d0:1e:f7:59:fd:04:1c:66:31:c6:a5:d7:52:a1:
    11:19:a2:b0:50:d2:18:e5:fe:a0:1f:4c:e7:69:21:
    1a:bf:7d:b4:9c:22:96:ee:1e:de:c7:04:3e:ce:76:
    cb:7a:b6:08:cc:59:10:57:84:56:e5:0d:f9:f9:e7:
    78:ba:ca:2b:29:aa:3a:17:4a:d4:ed:f2:1c:d2:f6:
    0c:52:41:de:bc:a7:2e:6f:3d:e7:90:09:b7:3c:80:
    fd:36:ac:4f:7a:89:ff:cb:8c:d7:6c:b4:ed:d0:d2:
    05:45:e0:a6:a7:fa:9f:2f:ce:09:f6:05:58:14:a8:
    3d:f5:d2:73:c0:35:8f:62:c0:35:2c:7c:16:cd:a6:
    52:03:f6:c2:ec:9b:87:33:28:8e:b5:7c:15:f5:b2:
    e2:0c:10:a4:32:7d:e5:75:47:c9:47:35:7b:10:77:
    8d:a0:75:de:8e:62:2c:e4:1f:5f:a2:a4:93:ea:36:
    c9:63:13:2a:66:99:6a:71:c0:3a:3b:30:15:27:89:
    2e:27:e0:22:69:d0:ff:d8:89:83:20:e6:f6:5a:bb:
    ba:31:f4:22:22:96:27:64:e1:f7:95:d4:dd:03:48:
    73:c5
publicExponent: 65537 (0x10001)
privateExponent:
    2e:49:23:15:cc:78:46:4b:bd:36:38:43:30:e2:6c:
    0c:7c:2b:f6:34:1e:12:1c:1b:de:a4:33:52:ff:8e:
    99:0b:a3:a0:86:c2:52:90:ca:04:67:c4:bd:81:f2:
    98:a3:e2:04:2e:ab:5b:91:76:10:ab:1e:9c:9d:1c:
    4d:a7:35:03:6e:63:a3:89:f1:3d:e4:2d:ae:d0:da:
    ae:7c:30:95:d6:10:cd:fa:e3:de:9d:40:3f:03:99:
    75:a7:7d:3b:cb:1d:8c:72:1b:a5:45:34:55:fc:3a:
    2e:0a:ae:fd:aa:53:be:c6:27:db:e8:08:6f:d0:39:
    65:bf:60:7a:ef:d5:4d:96:f7:42:c1:78:0d:d5:b1:
    40:93:93:e1:c4:66:df:55:3c:ef:2c:53:53:83:f3:
    1f:e7:1e:dd:16:13:1b:f5:51:47:e7:12:21:15:14:
    b4:3d:c9:96:df:b1:e9:8d:e9:f8:f5:ae:5a:aa:a9:
    87:d3:d1:be:3a:b5:5f:af:cd:0e:97:d6:3f:eb:39:
    fc:b9:fd:c1:2e:7a:5f:e0:a0:8d:39:1e:11:d0:cc:
    f4:55:84:4d:e7:f6:e7:1c:4b:76:45:7a:d2:57:db:
    69:ab:78:be:3a:de:7e:75:ae:8c:02:e4:c1:e6:5d:
    44:95:36:b3:17:f4:d6:37:24:30:2f:8e:33:42:75:
    a1
prime1:
    00:cf:bf:9b:7d:c6:62:40:db:13:cd:c1:24:61:a0:
    b0:20:b8:35:3d:f0:65:3a:0c:6d:79:91:31:5f:5c:
    4d:31:68:d4:10:38:b5:cf:ba:0f:e4:32:5a:eb:19:
    22:42:6a:57:1d:01:da:9e:a2:96:ae:3d:2b:b8:9f:
    fb:db:fc:91:21:57:ff:64:bf:7b:71:b8:9a:cb:2c:
    f7:09:77:92:f4:e0:c4:fb:f7:ad:42:ba:b1:b4:bb:
    d8:0d:76:08:bf:b1:3c:c8:a4:6f:fe:78:61:68:10:
    c9:c5:50:ae:ff:45:b3:f7:2c:af:65:c3:31:e3:c0:
    aa:9b:a6:cf:43:c5:d6:88:6d
prime2:
    00:ed:e3:dc:ab:b2:b0:d2:1d:55:56:11:c3:26:de:
    47:6f:88:1e:19:ea:88:ab:a4:8a:18:45:a5:b4:a8:
    e8:97:88:05:61:36:89:ef:bc:16:0e:1c:64:68:45:
    a0:1e:eb:f4:59:ef:40:06:80:ee:79:3d:89:82:92:
    a4:65:e1:5c:fe:36:06:a3:09:ef:f0:fe:b3:7f:00:
    cc:68:a5:65:78:fe:df:7c:4c:5a:89:f3:9e:83:0a:
    5c:d9:c0:7e:b9:51:f4:a6:86:f5:84:ac:cd:92:2f:
    c9:d3:85:05:d2:48:05:e6:d9:1b:d2:47:da:df:75:
    2a:23:b8:59:6a:77:cf:31:b9
exponent1:
    00:b2:72:dd:b7:6d:fe:db:c2:d3:01:fc:13:aa:20:
    82:e9:b8:11:53:c0:f8:d2:a1:f6:2f:64:3e:4a:b1:
    58:45:46:cc:51:9a:c4:e1:47:eb:7a:92:9e:72:2f:
    46:b4:51:ef:be:24:e7:8b:38:11:c2:5d:1c:ff:1b:
    bd:15:bb:9f:56:4a:e6:3f:be:12:e5:1e:33:28:ae:
    d9:0c:32:bf:f2:c4:57:b0:2f:91:8e:2e:1b:93:29:
    7f:6e:20:a3:67:35:71:06:cf:e8:bb:66:e0:14:a9:
    c7:07:4c:44:e5:55:14:3f:0f:0d:22:96:46:76:61:
    fa:aa:74:e0:b8:50:69:f5:09
exponent2:
    00:dd:31:56:f9:0b:30:a0:5e:07:ff:fc:8a:11:74:
    f2:c5:40:b2:84:74:6f:d9:f1:27:3a:44:60:3c:68:
    25:c4:e8:03:e0:8d:3e:9a:7b:af:a2:35:42:23:36:
    42:19:1e:5a:58:d9:3e:16:0e:02:2f:17:49:e9:f0:
    02:0e:7d:0a:e3:90:46:81:88:76:e3:05:60:cf:01:
    66:0f:13:63:13:ab:21:10:49:4a:92:24:c4:c9:a1:
    74:0e:40:8f:54:2f:61:25:0a:2d:f5:cc:5b:64:c7:
    2c:89:91:91:85:21:d4:e2:fe:12:81:66:b0:9b:c0:
    2c:8b:84:19:18:07:08:cf:79
coefficient:
    00:a6:08:ef:e8:98:ec:5c:6d:cc:1c:44:0c:f6:00:
    5d:fc:6d:cd:97:03:24:f9:8e:48:22:8c:19:08:09:
    cb:a8:b5:8a:ec:90:85:4c:f1:fd:dd:b8:ca:80:87:
    88:cb:b9:4e:72:82:45:cd:82:37:c2:77:70:8e:31:
    13:73:42:8c:df:ca:f2:a9:dc:b1:6f:2b:df:b4:d0:
    68:1f:70:ac:fa:a4:45:f1:02:90:1f:41:19:41:04:
    c7:3f:38:ad:65:b3:ea:ec:7a:95:4e:9d:0d:58:d6:
    c1:dc:23:8d:b0:a7:9b:73:b2:3d:48:2c:f3:e4:7c:
    e6:a8:b3:ff:48:4b:29:9a:c4

and

Client Random is

ede6b47287bcb60b56c7045f1672ea92e4e45aa5af6e902e359901ddaece4936

and

Server Random is

b3e805de8f3d430d2087824dec737840e0a98691fcaee2a2e7fb58d9c13fa3a7

and

Master Secret is

2494e0d48fa86f33855bdd7495e9e566da5e5f56e5507abecea9909af27814abca678306a2646baa6059c2a5aa05a369

and

target(That I want to decrypt) data is

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

I have tried to decrypt target data by this program;

import hashlib
from base64 import b64encode, b64decode

import cryptography
from cryptography.fernet import Fernet
master_secret = bytes.fromhex("2494e0d48fa86f33855bdd7495e9e566da5e5f56e5507abecea9909af27814abca678306a2646baa6059c2a5aa05a369")
label = b"key expansion"
hh1 = hashlib.sha1();
hh1.update(master_secret)
hh2 = hashlib.sha1();
hh2.update(label)
hh3 = hashlib.sha1();
tmptmptmp=bytes.fromhex("ede6b47287bcb60b56c7045f1672ea92e4e45aa5af6e902e359901ddaece4936"+"b3e805de8f3d430d2087824dec737840e0a98691fcaee2a2e7fb58d9c13fa3a7")
hh3.update(tmptmptmp)
last_key = hh1.hexdigest()+hh2.hexdigest()+hh3.hexdigest()
print(last_key)

#because base64encode of last_key's first 32byte is kE3/lhlz3PtJQZkus4AIrY0PA0yxRtIjBdoQa7RwjKo=
f = Fernet(b"kE3/lhlz3PtJQZkus4AIrY0PA0yxRtIjBdoQa7RwjKo=")
#because base64encode of target data is this
out = f.decrypt(b"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")
print(out)

But it does not work at all;

I think that it is wrong with getting key block, but I can't figure out correct specification from RFC 5246 6.3.;

Can someone please tell me the correct program in Python that can find the common key and decrypt the target data?

PS

I must have completely missed the fact that I needed to use the hmac-sha1 function instead of sha1. Perhaps that will work. If it works, I'll post the program here.