Problem:
Score : 400 points

Problem Statement

You are given positive integers N and M.

How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.

Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

2 6

Sample Output 1

4

Four sequences satisfy the condition: \{a_1, a_2\} = \{1, 6\}, \{2, 3\}, \{3, 2\} and \{6, 1\}.

Sample Input 2

3 12

Sample Output 2

18

Sample Input 3

100000 1000000000

Sample Output 3

957870001

Solution in Python (3.4.3) (judge verdict: Runtime Error)


iN,iM = [int(x) for x in input().split()]
iD = 10**9 +7 #法

def fPrimeCounter(iM):
    iS = int(iM ** 0.5) + 1
    aRet = []
    for i in range(2,iS + 1):
        iCounter = 0
        while iM % i == 0:
            iM = iM / i
            iCounter += 1
        if 0 < iCounter : aRet.append(iCounter)
        if iM == 1 : break
    if 1 < iM  : aRet.append(1)
    return aRet

#二分累乗法 iDを法として
def fBiPow(iX,iN,iD):
    iY = 1
    while iN > 0:
        if iN % 2 == 0:
            iX = iX * iX % iD
            iN = iN // 2
        else:
            iY = iX * iY % iD
            iN = iN - 1
    return iY

aP = fPrimeCounter(iM)
iMax = max(aP) + iN - 1
aM = [1] * (iMax + 1)
#階乗 iD を法とした
for i in range(1,iMax + 1):
    aM[i] = aM[i-1] * i  % iD
aInvM = [1]* (iMax + 1)
aInvM[iMax] = fBiPow(aM[iMax],iD-2,iD)
#階乗の逆元 iD を法とした
for i in range(iMax,0,-1):
    aInvM[i-1] = aInvM[i]*i % iD

iRet = 1
for i in aP:
    # i + iN -1  C  iN -1
    iRet *= aM[i+iN-1] * aInvM[iN-1] * aInvM[i] % iD
iRet %= iD
print(iRet)