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:

<image>

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 maxn = 1e6 + 10, Log = 23, mod = 1e9 + 7;
int dp[maxn][Log], par[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < maxn; i++) {
    if (par[i] == 0) {
      for (int j = i; j < maxn; j += i) par[j] = i;
    }
  }
  dp[0][0] = 1;
  for (int i = 1; i < Log; i++) {
    dp[0][i] = 2;
  }
  for (int r = 1; r < maxn; r++) {
    dp[r][0] = 1;
    for (int i = 1; i < Log; i++)
      dp[r][i] = (dp[r][i - 1] + dp[r - 1][i]) % mod;
  }
  int q;
  cin >> q;
  while (q--) {
    int r, a, ans = 1;
    cin >> r >> a;
    while (a > 1) {
      int p = par[a], c = 0;
      while (a % p == 0) c++, a /= p;
      ans = 1ll * ans * dp[r][c] % mod;
    }
    cout << ans << "\n";
  }
}