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 + 1;
const int mod = 1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
vector<int> p[N];
int f[N][20];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i < N; ++i)
    if (p[i].empty())
      for (int j = i; j < N; j += i) p[j].push_back(i);
  f[0][0] = 1;
  for (int i = 1; i < 20; ++i) f[0][i] = 2;
  for (int i = 1; i < N; ++i) {
    f[i][0] = 1;
    for (int j = 1; j < 20; ++j) f[i][j] = add(f[i][j - 1], f[i - 1][j]);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int r;
    cin >> r;
    int n;
    cin >> n;
    int ans = 1;
    for (int x : p[n]) {
      int e = 0;
      for (; n % x == 0; n /= x, ++e)
        ;
      ans = mul(ans, f[r][e]);
    }
    cout << ans << "\n";
  }
}