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 mod = (int)1e9 + 7;
const int N = 1000000;
int n, r, q;
int lp[N + 1];
int dp[21][N + 1];
int main() {
  for (int i = 2; i <= N; i += 2) lp[i] = 2;
  for (int i = 3; i <= 1000; i += 2)
    if (!lp[i]) {
      lp[i] = i;
      for (int j = i * i; j <= N; j += (i + i)) {
        lp[j] = i;
      }
    }
  for (int i = 2; i <= N; ++i) {
    if (!lp[i]) lp[i] = i;
  }
  for (int i = 1; i <= 20; ++i) dp[i][0] = 2;
  for (int j = 1; j <= N; ++j) {
    dp[0][j] = 1;
    for (int k = 1; k <= 20; ++k) {
      dp[k][j] = (dp[k - 1][j] + dp[k][j - 1]);
      if (dp[k][j] >= mod) dp[k][j] -= mod;
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &r, &n);
    long long ans = 1;
    for (int i = lp[n]; n != 1; i = lp[n]) {
      int p = 0;
      while (n % i == 0) {
        p++;
        n /= i;
      }
      ans = (ans * dp[p][r]) % mod;
    }
    printf("%d\n", ans);
  }
}