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 P = 1000000007, MAXN = 1000031, MAXL = 21;
bool ip[MAXN] = {1, 1};
int p[MAXN], l[MAXN] = {0, 1}, a[MAXN] = {0, 0};
void sieve() {
  for (int i = 2, c = 0; i < MAXN; i++) {
    if (!ip[i]) {
      p[c++] = i;
      l[i] = 1;
      a[i] = 1;
    }
    for (int j = 0; j < c && i * p[j] < MAXN; j++) {
      ip[i * p[j]] = 1;
      if (i % p[j]) {
        l[i * p[j]] = i;
        a[i * p[j]] = 1;
      } else {
        l[i * p[j]] = l[i];
        a[i * p[j]] = a[i] + 1;
        break;
      }
    }
  }
}
int f[MAXN][MAXL];
void pre() {
  std::fill(f[0], f[0] + MAXL, 2);
  f[0][0] = 1;
  for (int i = 1; i < MAXN; i++) {
    f[i][0] = 1;
    for (int j = 1; j < MAXL; j++) {
      f[i][j] = (f[i - 1][j] + f[i][j - 1]) % P;
    }
  }
}
int solve(int r, int n) {
  long long ans = 1;
  while (n > 1) {
    ans = ans * f[r][a[n]] % P;
    n = l[n];
  }
  return ans;
}
int main() {
  sieve();
  pre();
  int q = 0;
  scanf("%d", &q);
  while (q--) {
    int r = 0, n = 0;
    scanf("%d%d", &r, &n);
    printf("%d\n", solve(r, n));
  }
  return 0;
}