Problem: Bash got tired on his journey to become the greatest Pokemon master. So he decides to take a break and play with functions.

Bash defines a function f0(n), which denotes the number of ways of factoring n into two factors p and q such that gcd(p, q) = 1. In other words, f0(n) is the number of ordered pairs of positive integers (p, q) such that p·q = n and gcd(p, q) = 1.

But Bash felt that it was too easy to calculate this function. So he defined a series of functions, where fr + 1 is defined as:

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Where (u, v) is any ordered pair of positive integers, they need not to be co-prime.

Now Bash wants to know the value of fr(n) for different r and n. Since the value could be huge, he would like to know the value modulo 109 + 7. Help him!

Input

The first line contains an integer q (1 ≤ q ≤ 106) — the number of values Bash wants to know.

Each of the next q lines contain two integers r and n (0 ≤ r ≤ 106, 1 ≤ n ≤ 106), which denote Bash wants to know the value fr(n).

Output

Print q integers. For each pair of r and n given, print fr(n) modulo 109 + 7 on a separate line.

Example

Input

5
0 30
1 25
3 65
2 5
4 48


Output

8
5
25
4
630

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
int c[22][1001000], p[1001000], v[40];
int main() {
  int i, j;
  for (i = 0; i < 1001000; c[0][i] = 1, i++)
    ;
  for (i = 1; i < 22; i++)
    for (c[i][0] = 2, j = 1; j < 1001000;
         c[i][j] = (c[i - 1][j] + c[i][j - 1]) % 1000000007, j++)
      ;
  for (i = 2; i * i < 1001000; i++)
    if (!p[i])
      for (j = i * i; j < 1001000; p[j] = i, j += i)
        ;
  for (i = 2; i < 1001000; i++)
    if (!p[i]) p[i] = i;
  int q;
  scanf("%d", &q);
  for (; q--;) {
    int l, n, r;
    scanf("%d%d", &r, &n);
    l = 0;
    for (; n > 1; v[l++] = p[n], n /= p[n])
      ;
    sort(v, v + l);
    int z = 1;
    for (i = 0; i < l; z = (long long)z * c[j][r] % 1000000007, i += j)
      for (j = 1; i + j < l && v[i] == v[i + j]; j++)
        ;
    printf("%d\n", z);
  }
  return 0;
}