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;
int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
int mult(int x, int y) { return ((long long)x * y) % MOD; }
const int N = (int)1e6 + 3;
const int K = 20;
int d[N];
int dp[N][K];
int main() {
  for (int x = 2; x < N; x++) d[x] = -1;
  for (int x = 2; x < N; x++) {
    if (d[x] != -1) continue;
    for (int y = x; y < N; y += x)
      if (d[y] == -1) d[y] = x;
  }
  dp[0][0] = 1;
  for (int i = 1; i < K; i++) dp[0][i] = 2;
  for (int i = 1; i < N; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < K; j++) dp[i][j] = add(dp[i - 1][j], dp[i][j - 1]);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int ans = 1;
    while (y > 1) {
      int p = d[y];
      int t = 0;
      while (y % p == 0) {
        t++;
        y /= p;
      }
      ans = mult(ans, dp[x][t]);
    }
    printf("%d\n", ans);
  }
  return 0;
}