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 mod = 1e9 + 7;
long long dp[1000006][22];
int lowp[1000006];
void pre() {
  for (int j = 1; j < 22; j++) dp[0][j] = 2;
  for (int i = 0; i < 1000006; i++) dp[i][0] = 1;
  for (int i = 1; i < 1000006; i++)
    for (int j = 1; j < 22; j++) {
      dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
      if (dp[i][j] >= mod) dp[i][j] -= mod;
    }
  for (int i = 2; i < 1000006; i++) lowp[i] = i;
  for (int i = 2; i < 1000006; i++) {
    if (lowp[i] == i) {
      for (int j = i + i; j < 1000006; j += i) {
        if (lowp[j] == j) {
          lowp[j] = i;
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  pre();
  int q;
  cin >> q;
  while (q--) {
    int r, n;
    cin >> r >> n;
    long long ans = 1;
    while (n > 1) {
      int p = lowp[n];
      int k = 0;
      while (p * (n / p) == n) {
        n /= p;
        k++;
      }
      ans *= dp[r][k];
      ans %= mod;
    }
    cout << ans << "\n";
  }
  return 0;
}