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;
using ll = long long;
using ld = long double;
const int N = 1e6 + 6;
const int L = 20;
const int mod = 1e9 + 7;
int c[N][L];
vector<int> pr, md;
void getPrimes(int n) {
  md.assign(n + 1, 0);
  pr.clear();
  for (int i = 2; i <= n; ++i) {
    if (!md[i]) pr.push_back(md[i] = i);
    for (int p : pr) {
      if (p > md[i] || 1LL * p * i > n) break;
      md[p * i] = p;
    }
  }
}
int main() {
  for (int i = 0; i < N; ++i) c[i][0] = 1;
  for (int i = 1; i < L; ++i) c[0][i] = 2;
  for (int i = 1; i < N; ++i)
    for (int j = 1; j < L; ++j) {
      c[i][j] = c[i - 1][j] + c[i][j - 1];
      if (c[i][j] >= mod) c[i][j] -= mod;
    }
  getPrimes(N);
  int m;
  scanf("%d", &m);
  while (m--) {
    int r, n;
    scanf("%d%d", &r, &n);
    ll ans = 1;
    for (int x = n; x > 1;) {
      int p = md[x], q = 0;
      while (x % p == 0) {
        x /= p;
        ++q;
      }
      ans = ans * c[r][q] % mod;
    }
    printf("%d\n", (int)ans);
  }
  return 0;
}