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;
int pr[1000005];
int prf[1000001][21];
void SOE(int N) {
  for (int p = 2; p <= N; p++) {
    if (!pr[p]) {
      pr[p] = p;
      long long ch = ((long long)(p) * (long long)(p));
      if (ch > N) continue;
      for (int i = p * p; i <= N; i += p) pr[i] = p;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  SOE(1000000);
  for (int r = 0; r <= 1000000; ++r) prf[r][0] = 1LL;
  for (int i = 1; i <= 20; ++i) prf[0][i] = prf[0][i - 1] + 2LL;
  for (int i = 1; i <= 20; ++i)
    for (int r = 1; r <= 1000000; ++r) {
      prf[r][i] = (prf[r][i - 1] + prf[r - 1][i]) % 1000000007;
    }
  int q;
  cin >> q;
  while (q--) {
    int r, n;
    cin >> r >> n;
    if (n == 1) {
      cout << 1 << '\n';
      continue;
    }
    vector<long long> tem;
    int y = n;
    long long ans = 1LL;
    while (y > 1) {
      int v = pr[y];
      long long ct = 0;
      while (y % v == 0) {
        ct++;
        y /= v;
      }
      ans = (ans * ((prf[r][ct] - prf[r][ct - 1]) % 1000000007 + 1000000007) %
             1000000007) %
            1000000007;
    }
    cout << ans << '\n';
  }
}