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>
const int MAXN = 1E6 + 10, MAXD = 20, MAXR = 1E6 + 10;
const int MOD = 1E9 + 7;
int np, prime[MAXN];
int isnp[MAXN], md[MAXN];
int f[MAXR][MAXD];
void getPrime(int n = MAXN - 1) {
  np = 0;
  isnp[0] = isnp[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!isnp[i]) prime[np++] = md[i] = i;
    for (int t = n / i, j = 0; prime[j] <= t; ++j) {
      isnp[prime[j] * i] = 1;
      md[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  getPrime();
  for (int i = 0; i < MAXD; ++i) f[0][i] = 1 + !!i;
  for (int i = 1; i < MAXR; ++i) {
    f[i][0] = 1;
    for (int j = 1; j < MAXD; ++j)
      if ((f[i][j] = f[i][j - 1] + f[i - 1][j]) >= MOD) f[i][j] -= MOD;
  }
  int q, r, n;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    int ans = 1;
    for (; n > 1;) {
      int d = md[n], cnt = 0;
      for (; md[n] == d; ++cnt, n /= md[n])
        ;
      ans = 1ll * ans * f[r][cnt] % MOD;
    }
    printf("%d\n", ans);
  }
  return 0;
}