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 N = 1e6 + 6;
const int MO = 1e9 + 7;
int dp[22][N], f[22][N];
void Add(int &x, int y) {
  x += y;
  if (x >= MO) x -= MO;
}
void init() {
  for (int j = 0; j < N; ++j) {
    for (int i = 0; i <= 20; ++i) {
      if (j == 0) {
        dp[i][j] = i == 0 ? 1 : 2;
      } else {
        dp[i][j] = i == 0 ? 1 : f[i][j - 1];
      }
      for (int k = 0; k <= i; ++k) {
        Add(f[i][j], dp[k][j]);
      }
    }
  }
}
namespace PM {
static constexpr int N = 1e3 + 5;
static constexpr int MO = 1e9 + 7;
using array = int[N];
array mark, prime, mobius;
int count = 0;
std::vector<int> factor[N];
void GetPrime() {
  factor[1].emplace_back(1);
  for (int i = 2; i < N; ++i) {
    if (!mark[i]) {
      prime[count++] = i;
      mobius[i] = MO - 1;
      factor[i].emplace_back(i);
    }
    for (int j = 0; j < count && i * prime[j] < N; ++j) {
      int now = i * prime[j];
      mark[now] = prime[j];
      mobius[now] = (MO - mobius[i]) % MO;
      factor[now] = factor[i];
      factor[now].emplace_back(prime[j]);
      if (i % prime[j] == 0) {
        mobius[now] = 0;
        factor[now].pop_back();
        break;
      }
    }
  }
}
}  // namespace PM
int main() {
  init();
  int q;
  scanf("%d", &q);
  PM::GetPrime();
  while (q--) {
    int r, c;
    scanf("%d%d", &r, &c);
    int ans = 1;
    for (int i = 0; i < PM::count && PM::prime[i] * PM::prime[i] <= c; ++i) {
      if (c % PM::prime[i] == 0) {
        int cnt = 0;
        while (c % PM::prime[i] == 0) {
          c /= PM::prime[i];
          cnt++;
        }
        ans = 1LL * ans * dp[cnt][r] % MO;
      }
    }
    if (c > 1) ans = 1LL * ans * dp[1][r] % MO;
    printf("%d\n", ans);
  }
  return 0;
}