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 N = 1e6 + 7;
const int M = 1e3 + 7;
const int MOD = 1e9 + 7;
int p[M], vis[M], len;
void prime() {
  for (int i = 2; i < M; ++i) {
    if (!vis[i]) {
      p[len++] = i;
    }
    for (int j = 0; j < len && i * p[j] < M; ++j) {
      vis[i * p[j]] = true;
      if (!(i % p[j])) break;
    }
  }
}
int val[N][20];
int main() {
  int q, r, n;
  prime();
  for (int i = 0; i < 20; ++i) {
    val[0][i] = i == 0 ? 1 : 2;
  }
  for (int i = 1; i < N; ++i) {
    for (int j = 0; j < 20; ++j) {
      val[i][j] = j == 0 ? 1 : val[i - 1][j] + val[i][j - 1];
      if (val[i][j] >= MOD) val[i][j] -= MOD;
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    long long ans = 1;
    for (int i = 0; i < len; ++i) {
      int num = 0;
      while (n % p[i] == 0) {
        n /= p[i];
        num++;
      }
      ans *= val[r][num];
      ans %= MOD;
      if (n == 1) break;
    }
    if (n > 1) {
      ans *= val[r][1];
      ans %= MOD;
    }
    printf("%lld\n", ans);
  }
  return 0;
}