exception raised when decrypt with private key re-constructed by only n, d

the following code would raise the exception
python 3.5 64bit on windows 10 64bit

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5 as cipher_pkcs1

    public_key = """-----BEGIN PUBLIC KEY-----
MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCIk50gdmuHTUqE1Iet3f3/uDg3/vLNpQlCoFTs
G+NYvQAToihP+HSxAg5lnzYvfruNV0QnIPRwSKFbPmNeHnMmPXogmjyJB1hjQcKEc/SvErMoMFuA
2phryr+gT1kG0s6MtLN1Y4LVr2tNiqFLDYr7WCn52Zj842PWSiGu/elx4QIDAQAB
-----END PUBLIC KEY-----"""

# original private key which contains n,d,e,p,q
# private_key = """-----BEGIN PRIVATE KEY-----
# MIICdgIBADANBgkqhkiG9w0BAQEFAASCAmAwggJcAgEAAoGBAIiTnSB2a4dNSoTUh63d/f+4ODf+
# 8s2lCUKgVOwb41i9ABOiKE/4dLECDmWfNi9+u41XRCcg9HBIoVs+Y14ecyY9eiCaPIkHWGNBwoRz
# 9K8SsygwW4DamGvKv6BPWQbSzoy0s3VjgtWva02KoUsNivtYKfnZmPzjY9ZKIa796XHhAgMBAAEC
# gYBq7qhnclBebzWZKo3Qo67rOOFLdmHEgSLu93mGI4+tOs2EkcHeWzmCHYoIJs/V//P4hgrmV/cK
# stMjDYXfexfL+mzRUxQg5XBfiuqEahRTxOBxvXYnqJlyeNVtsxatF+bjA8m3ZZ5nuG+41ETHzkfS
# LAQsCzkS+IQP/Kr4POVVwQJBAPHQBfB0XsT8mtZrk6GtqSsDli/+/GqRgPP6jBvIY0yG6lwtBFW4
# O/NDmKrtSvATzuKF+3tWlmP5cRUj+xNiBE8CQQCQlvee0ZCMMM/d1r6WmrDYVpyg7kDAfEkFghKk
# ao+ntPDCyf93NjfqLV4p33HuSmOt9zYW3rUeJzhC/FJAHSrPAkAi+24wiBVBuC2KQYvdAuuLPnyW
# uvO3Uw9tvFxSlJ3Q71+uMBEFeZ6qpafLaoLB0ovGd42XHUKqeeajIU5sUmy/AkAWCk04sm08UXx+
# Lmvi/TWT5K//gBHyxX9EDa3VzkCTqetBZhVBEMy5L6046MyvLDvBmvttxQkKzaOv4lKPX2LVAkEA
# 64qQuwJD+dLLMOpL34fRZk1NXzqi95S0G2aseGphJ8xI37QXHV/ogZ6y0N1WHNcnc6L/hXVlYrH+
# OOLi6VEyjw==
# -----END PRIVATE KEY-----"""

    #private key re-constructed by n d and p q not present
    private_key = """-----BEGIN PRIVATE KEY-----
MIIBNgIBADANBgkqhkiG9w0BAQEFAASCASAwggEcAgEAAoGBAIiTnSB2a4dNSoTUh63d/f+4ODf+
8s2lCUKgVOwb41i9ABOiKE/4dLECDmWfNi9+u41XRCcg9HBIoVs+Y14ecyY9eiCaPIkHWGNBwoRz
9K8SsygwW4DamGvKv6BPWQbSzoy0s3VjgtWva02KoUsNivtYKfnZmPzjY9ZKIa796XHhAgEAAoGA
au6oZ3JQXm81mSqN0KOu6zjhS3ZhxIEi7vd5hiOPrTrNhJHB3ls5gh2KCCbP1f/z+IYK5lf3CrLT
Iw2F33sXy/ps0VMUIOVwX4rqhGoUU8Tgcb12J6iZcnjVbbMWrRfm4wPJt2WeZ7hvuNREx85H0iwE
LAs5EviED/yq+DzlVcECAQACAQACAQACAQACAQA=
-----END PRIVATE KEY-----"""

    message = 'abcdef'.encode()
    pub_key = RSA.importKey(public_key)
    cipher = cipher_pkcs1.new(pub_key)
    encrypted = cipher.encrypt(message)
    print("{} encrypted as {}".format(message, encrypted))

    pri_key = RSA.importKey(private_key)
    cipher = cipher_pkcs1.new(pri_key)
    sentinel = None
    decrypted = cipher.decrypt(encrypted, sentinel)
    print("{} decrypted as {}".format(encrypted, decrypted))

the error looks like:
File "C:\Anaconda3\lib\site-packages\Crypto\Cipher\PKCS1_v1_5.py", line 206, in decrypt
    m = self._key.decrypt(ct)
  File "C:\Anaconda3\lib\site-packages\Crypto\PublicKey\RSA.py", line 174, in decrypt
    return pubkey.pubkey.decrypt(self, ciphertext)
  File "C:\Anaconda3\lib\site-packages\Crypto\PublicKey\pubkey.py", line 93, in decrypt
    plaintext=self._decrypt(ciphertext)
  File "C:\Anaconda3\lib\site-packages\Crypto\PublicKey\RSA.py", line 240, in _decrypt
    mp = self.key._decrypt(cp)
  File "C:\Anaconda3\lib\site-packages\Crypto\PublicKey\_slowmath.py", line 54, in _decrypt
    m1 = pow(c, self.d % (self.p-1), self.p)
ValueError: pow() 3rd argument cannot be 0

using the original private key instead of the re-constructed one, everything is fine.
with the following modifications, the decryption would work:
1. _slowmath.py line 53
-if (hasattr(self,'p') and hasattr(self,'q') and hasattr(self,'u')):
+if (hasattr(self,'p') and hasattr(self,'q') and hasattr(self,'u') and self.p > 0 and self.q > 0):
2. RSA.py line 235
-r = getRandomRange(1, self.key.n-1, randfunc=self._randfunc)
+r = 1

have no idea why the r has to be 1, 1 just works.