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>
using namespace std;
const int N = 1000005;
const int Q = 1e9 + 7;
int fac[N];
bool f[N];
int c[N][22];
int main() {
  for (int i = 2; i < N; ++i) {
    if (!f[i]) {
      fac[i] = i;
      for (int j = i + i; j < N; j += i) {
        if (!fac[j]) fac[j] = i;
        f[j] = 1;
      }
    }
  }
  c[0][0] = 1;
  for (int i = 1; i < 22; ++i) {
    c[0][i] = 2;
  }
  for (int i = 1; i < N; ++i) {
    c[i][0] = c[i - 1][0];
    for (int j = 1; j < 22; ++j) {
      c[i][j] = (c[i - 1][j] + c[i][j - 1]) % Q;
    }
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int res = 1;
    while (n > 1) {
      int x = fac[n], y = 0, v = 1;
      while (n % x == 0) {
        n /= x;
        y++;
      }
      res = (long long)res * c[r][y] % Q;
    }
    printf("%d\n", res);
  }
}