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 long long N = 1e6 + 2;
const long long X = 1e6;
const long long mod = 1e9 + 7;
long long dp[N][22], divi[N];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, i, j, k, l, q, ans = 1;
  for (i = 1; i <= 20; i++) {
    dp[0][i] = 2;
  }
  dp[0][0] = 1;
  for (i = 1; i <= X; i++) {
    k = 0;
    for (j = 0; j <= 20; j++) {
      k += dp[i - 1][j];
      if (k >= mod) {
        k -= mod;
      }
      dp[i][j] = k;
    }
  }
  for (i = 2; i <= X; i++) {
    if (!divi[i]) {
      for (j = i; j <= X; j += i) {
        if (!divi[j]) {
          divi[j] = i;
        }
      }
    }
  }
  cin >> q;
  while (q--) {
    ans = 1;
    cin >> k >> n;
    while (n > 1) {
      j = divi[n];
      i = 0;
      while (n % j == 0) {
        n /= j;
        i++;
      }
      ans *= dp[k][i];
      ans %= mod;
    }
    cout << ans << '\n';
  }
}