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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
int T;
long long dp[22][1000005];
const long long MOD = 1e9 + 7;
int pr[1000005];
bool sieve[1000005];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  for (int i = 2; i <= 1000000; i++) {
    if (!sieve[i]) {
      for (int j = i; j <= 1000000; j += i) {
        sieve[j] = 1;
        pr[j] = i;
      }
    }
  }
  fill(dp[0], dp[0] + 1000001, 1);
  for (int p = 1; p <= 20; p++) {
    dp[p][0] = 2;
    for (int n = 1; n <= 1000000; n++) {
      dp[p][n] = (dp[p - 1][n] + dp[p][n - 1]) % MOD;
    }
  }
  while (T--) {
    long long r, n;
    cin >> r >> n;
    long long ans = 1;
    while (n - 1) {
      int p = pr[n];
      int cnt = 0;
      while (n % p == 0) {
        n /= p;
        cnt++;
      }
      ans *= dp[cnt][r];
      ans %= MOD;
    }
    cout << ans << "\n";
  }
}