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:

<image>

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 maxx = 1e6 + 2, mxt = 20, mod = 1e9 + 7;
int n, ps[maxx][mxt];
bool ip[maxx];
vector<int> v[maxx];
void era() {
  for (int i = 2; i < maxx; i++) {
    if (!ip[i]) {
      v[i].push_back(i);
      for (int j = 2 * i; j < maxx; j += i) ip[j] = true, v[j].push_back(i);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  era();
  ps[0][0] = 1;
  for (int i = 1; i < mxt; i++) ps[0][i] = 2;
  for (int i = 1; i < maxx; i++) {
    ps[i][0] = 1;
    for (int j = 1; j < mxt; j++)
      ps[i][j] = (ps[i][j - 1] + ps[i - 1][j]) % mod;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int nn, r, n, p = 0;
    long long ans = 1ll;
    cin >> r >> n;
    nn = n;
    while (n > 1) {
      int t = 0, x = v[nn][p];
      while (n > 1 && n % x == 0) n /= x, t++;
      ans = (ans * ps[r][t]) % mod;
      p++;
    }
    cout << ans << '\n';
  }
  return 0;
}