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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int mod = 1e9 + 7;
const int mxN = 1e6 + 10;
const int C = 20;
int g[mxN][C];
int pref[mxN][C];
int minDiv[mxN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  for (int i = 0; i < C; i++) {
    g[0][i] = (i > 0) + 1;
    pref[0][i] = pref[0][max(0, i - 1)] + g[0][i];
  }
  for (int i = 1; i < mxN; i++) {
    minDiv[i] = i;
    for (int j = 0; j < C; j++) {
      g[i][j] = pref[i - 1][j];
      pref[i][j] = (pref[i][max(0, j - 1)] + g[i][j]) % mod;
    }
  }
  for (int i = 2; i < mxN; i++) {
    if (minDiv[i] != i) continue;
    for (int j = i * 2; j < mxN; j += i) minDiv[j] = min(minDiv[j], i);
  }
  int q;
  cin >> q;
  while (q--) {
    int r, n;
    cin >> r >> n;
    int ans = 1;
    while (n > 1) {
      int d = minDiv[n];
      int cnt = 0;
      while (n % d == 0) {
        n /= d;
        cnt++;
      }
      ans = 1ll * ans * g[r][cnt] % mod;
    }
    cout << ans << '\n';
  }
  return 0;
}