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 int maxn = 1e6 + 5, mod = 1e9 + 7, P = 20;
int f[maxn][P], s[maxn], n, r, q;
long long ans;
void seive() {
  for (int i = 2; i < maxn; i++)
    if (!s[i]) {
      for (int j = i; j < maxn; j += i) {
        s[j] = i;
      }
    }
}
void init() {
  for (int i = 0; i < maxn; i++) f[i][0] = 1;
  for (int i = 1; i < P; i++) f[0][i] = 2;
  for (int i = 1; i < maxn; i++) {
    for (int j = 1; j < P; j++) {
      f[i][j] = (f[i][j - 1] + f[i - 1][j]) % mod;
    }
  }
}
void solve() {
  long long ans = 1;
  while (n != 1) {
    int uoc = s[n], cnt = 0;
    while (n % uoc == 0) {
      n = n / uoc;
      cnt++;
    }
    ans = ans * f[r][cnt] % mod;
  }
  printf("%I64d\n", ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  seive();
  init();
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    solve();
  }
  return 0;
}