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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	inv,now,n,m,ans,mo,p,facnm:int64;
	fac:array[0..30]of int64;
	i,j:Longint;
function power(a,b:int64):int64;
var ret:int64;
begin
	if b=0 then power:=1
	else begin
		ret:=power(a*a mod 1000000007,b div 2);
		if b mod 2=1 then ret:=ret*a mod 1000000007;
		power:=ret;
	end;
end;
begin
	read(n,m);
	ans:=1;
	mo:=1000000007;
	fac[0]:=1;
	for i:=1 to 30 do fac[i]:=fac[i-1]*i mod mo;
	facnm:=1;
	for i:=2 to n-1 do facnm:=facnm*i mod mo;
	inv:=power(facnm,mo-2);
	i:=2;
	while i*i<=m do begin
		if m mod i=0 then begin
			p:=0;
			while m mod i=0 do begin
				m:=m div i;
				inc(p);
			end;
			now:=facnm;
			for j:=1 to p do now:=now*(n+j-1)mod mo;
			ans:=ans*now mod mo*power(fac[p],mo-2)mod mo*inv mod mo;
		end;
		inc(i);
	end;
	if m>1 then ans:=ans*facnm mod mo*n mod mo*inv mod mo;
	writeln(ans);
end.