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 MAXN = (1 << 20);
const int MAXLOG = 22;
const int mod = (int)1e9 + 7;
const int mod_inv2 = mod / 2 + 1;
int q;
void read() { cin >> q; }
int dp[MAXN][MAXLOG];
int lp[MAXN];
void precompute() {
  for (int i = 2; i < MAXN; i++)
    if (!lp[i])
      for (int j = i; j < MAXN; j += i)
        if (!lp[j]) lp[j] = i;
  dp[0][0] = 1;
  for (int r = 1; r < MAXLOG; r++) dp[0][r] = 2;
  for (int i = 1; i < MAXN; i++)
    for (int r = 0; r < MAXLOG; r++)
      dp[i][r] = (dp[i - 1][r] + (r ? dp[i][r - 1] : 0)) % mod;
}
void solve() {
  precompute();
  for (int t = 0; t < q; t++) {
    int n, r;
    cin >> r >> n;
    int64_t res = 1;
    while (n != 1) {
      int cnt = 0, div = lp[n];
      while (n % div == 0) cnt++, n /= div;
      res = (res * 1ll * dp[r][cnt]) % mod;
    }
    cout << res << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}