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>
#pragma comment(linker, "/STACK: 1024000000,1024000000")
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 1e6 + 16;
int f[N][22], pri[N], ct = 0;
bool vis[N];
void init() {
  for (int i = 2; i < N; i++) {
    if (!vis[i]) pri[ct++] = i;
    for (int j = 0; j < ct && 1ll * pri[j] * i < N; j++) {
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
  f[0][0] = 1;
  for (int i = 1; i < 22; i++) f[0][i] = 2;
  for (int sm = 0, i = 1; i < N; i++) {
    for (int j = 0; j < 22; j++) {
      sm = (int)((1ll * sm + f[i - 1][j]) % MOD);
      f[i][j] = (int)((1ll * f[i][j] + sm) % MOD);
    }
    sm = 0;
  }
}
int main() {
  int r, n;
  init();
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &r, &n);
    int ans = 1;
    for (int i = 0; i < ct && pri[i] * pri[i] <= n; i++) {
      if (n % pri[i] == 0) {
        int mi = 0;
        while (n % pri[i] == 0) {
          mi++;
          n /= pri[i];
        }
        ans = (int)((1ll * ans * f[r][mi]) % MOD);
      }
    }
    if (n > 1) ans = (int)((1ll * ans * f[r][1]) % MOD);
    printf("%d\n", ans % MOD);
  }
  return 0;
}