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:

<image>

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>
using namespace std;
const int MOD = 1000000007;
const int N = 1000000;
int prime[N + 5];
int f[N + 5][25];
int q, r, c;
int main() {
  for (int i = 2; i <= N; i++) {
    for (int j = i; j <= N; j += i) {
      if (prime[j] == 0) {
        prime[j] = i;
      }
    }
  }
  f[0][0] = 1;
  for (int i = 1; i <= 22; i++) {
    f[0][i] = 2;
  }
  for (int i = 1; i <= N; i++) {
    f[i][0] = 1;
    for (int j = 1; j <= 22; j++) {
      f[i][j] = (f[i][j - 1] + f[i - 1][j]) % MOD;
    }
  }
  scanf("%d", &q);
  while (q--) {
    int ans = 1;
    scanf("%d %d", &r, &c);
    while (c != 1) {
      int s = 0, x = prime[c];
      while (c % x == 0) {
        c /= x;
        s++;
      }
      ans = (long long)ans * f[r][s] % MOD;
    }
    printf("%d\n", ans);
  }
  return 0;
}