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>
const int N = 1000010;
const int M = 30;
const int moder = 1e9 + 7;
int pre[N][M], min[N];
std::vector<int> prime;
int main() {
  for (int i = 2; i < N; ++i) {
    if (!min[i]) {
      prime.push_back(i);
      min[i] = i;
    }
    for (int j = 0, k, sz = (int)prime.size(); j < sz && i * prime[j] < N;
         ++j) {
      min[k = i * prime[j]] = prime[j];
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
  pre[0][0] = 1;
  for (int i = 1; i < M; ++i) {
    pre[0][i] = 2;
  }
  for (int i = 1; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      pre[i][j] = ((j ? pre[i][j - 1] : 0) + pre[i - 1][j]) % moder;
    }
  }
  int test;
  scanf("%d", &test);
  while (test--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int ans = 1;
    while (n > 1) {
      int x = min[n], cnt = 0;
      while (!(n % x)) {
        n /= x;
        ++cnt;
      }
      ans = 1ll * ans * pre[r][cnt] % moder;
    }
    printf("%d\n", ans);
  }
  return 0;
}