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 M = 1000050, N = M, mod = (int)1e9 + 7;
int p[M];
int mu[M], pm[M], pms;
void get_mobius_and_sieve() {
  mu[1] = 1;
  for (int i = 1; i < M; i++) p[i] = i;
  pms = 0;
  for (int i = 2; i < M; i++) {
    if (p[i] == i) {
      pm[pms++] = i;
      mu[i] = -1;
    }
    for (int j = 0; j < pms && i * pm[j] < M; j++) {
      p[i * pm[j]] = pm[j];
      if (i % pm[j] == 0) {
        mu[i * pm[j]] = 0;
        break;
      }
      mu[i * pm[j]] = -mu[i];
    }
  }
}
int T;
long long pw[N][20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  get_mobius_and_sieve();
  for (int r = 0; r < N; r++) {
    pw[r][0] = 1;
    if (r == 0)
      fill(pw[r], pw[r] + N, 2);
    else
      for (int j = 1; j < 20; j++)
        pw[r][j] = (pw[r][j - 1] + (r == 0 ? 0 : pw[r - 1][j])) % mod;
  }
  cin >> T;
  while (T--) {
    int r, n;
    long long res = 1;
    cin >> r >> n;
    while (n > 1) {
      int cnt = 0;
      int cur_p = p[n];
      while (n % cur_p == 0) n /= cur_p, cnt++;
      res = (res * pw[r][cnt]) % mod;
    }
    cout << res << "\n";
  }
}