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 long long N = 1e6 + 5;
const long long mod = 1e9 + 7;
vector<long long> prime(N, 0);
const long long P = 25;
vector<vector<long long> > dp(N, vector<long long>(P, 0));
void sieve() {
  prime[1] = 1;
  for (long long i = 2; i < N; i++) {
    if (prime[i] == 0) {
      for (long long j = i; j < N; j += i) prime[j] = i;
    }
  }
}
void calc() {
  for (long long i = 0; i < N; i++) dp[i][0] = 1;
  for (long long i = 1; i < P; i++) dp[0][i] = 2;
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < P; j++)
      dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  sieve();
  calc();
  long long q;
  cin >> q;
  while (q-- > 0) {
    long long r, n;
    cin >> r >> n;
    long long ans = 1;
    while (n != 1) {
      long long cnt = 0, p = prime[n];
      while (n % p == 0) n /= p, cnt++;
      ans = (ans * dp[r][cnt]) % mod;
    }
    cout << ans << "\n";
  }
}