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: Time Limit Exceeded)

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

def fPrimeCounter(iM):
    i = 2
    iS = int(iM ** 0.5) + 1
    aRet = []
    while i < iS or iM != 1:
        iCounter = 0
        while iM % i == 0:
            iM = iM // i
            iCounter += 1
        if 0 < iCounter : aRet.append(iCounter)
        i += 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)
if aP == [] :
    print (1)
else:
    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)