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 maxn = 1000000 + 10;
const int md = 1e9 + 7;
int f[maxn][22], q, p[maxn], tot, check[maxn], sum[maxn][22], ntr[maxn];
void pre() {
  for (int i = 2; i <= 1000000; i++) {
    if (!check[i]) {
      p[++tot] = i;
      ntr[i] = i;
    }
    for (int j = 1; j <= tot && (long long)i * p[j] <= 1000000; j++) {
      check[i * p[j]] = 1;
      ntr[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  f[0][0] = 1;
  for (int i = 1; i <= 20; i++) f[0][i] = 2;
  sum[0][0] = f[0][0];
  for (int i = 1; i <= 20; i++) sum[0][i] = (f[0][i] + sum[0][i - 1]) % md;
  for (int i = 1; i <= 1000000; i++) {
    for (int j = 0; j <= 20; j++) f[i][j] = sum[i - 1][j];
    sum[i][0] = f[i][0];
    for (int j = 1; j <= 20; j++) sum[i][j] = (f[i][j] + sum[i][j - 1]) % md;
  }
}
int main() {
  scanf("%d", &q);
  pre();
  for (int i = 1; i <= q; i++) {
    int n, r;
    scanf("%d%d", &r, &n);
    long long ans = 1;
    int t = n;
    while (t > 1) {
      int cnt = 0, hm = ntr[t];
      while (t % hm == 0) {
        cnt++;
        t /= hm;
      }
      ans = ans * f[r][cnt] % md;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}