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;
int p[1000000];
int d[20][1000001];
int main() {
  int i, j, m = 0, r, n, T;
  int ans;
  vector<int> v;
  for (i = 2; i < 1000; i++)
    if (!p[i])
      for (j = i * i; j < 1000000; j += i) p[j] = 1;
  for (i = 2; i < 1000000; i++)
    if (!p[i]) p[m++] = i;
  for (i = 0; i < 20; i++) d[i][0] = 2;
  d[0][0] = 1;
  for (i = 1; i <= 1000000; i++) {
    d[0][i] = 1;
    for (j = 1; j < 20; j++) d[j][i] = (d[j][i - 1] + d[j - 1][i]) % 1000000007;
    ;
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &r, &n);
    v.clear();
    for (i = 0; i < m && p[i] * p[i] <= n; i++)
      if (n % p[i] == 0) {
        j = 0;
        while (n % p[i] == 0) {
          n /= p[i];
          j++;
        }
        v.push_back(j);
      }
    if (n != 1) v.push_back(1);
    ans = 1;
    for (i = 0; i < v.size(); i++)
      ans = (long long)ans * d[v[i]][r] % 1000000007;
    ;
    printf("%d\n", ans);
  }
  return 0;
}