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;
vector<int> facs[1000009];
int dp[1000002][20], inv, mod = 1000000007;
int main() {
  for (int i = 0; i <= 20; i++) dp[0][i] = 2;
  dp[0][0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    int pref = 0;
    for (int j = 0; j <= 20; j++) {
      pref = (pref + dp[i - 1][j]) % mod;
      dp[i][j] = pref;
    }
  }
  for (int i = 2; i <= 1000000; i++)
    if (!facs[i].size()) {
      for (int j = i; j <= 1000000; j += i) {
        facs[j].push_back(i);
      }
    }
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d %d", &r, &n);
    int sol = 1;
    for (int i = 0; i < facs[n].size(); i++) {
      int c = n, x = 0;
      while (c % facs[n][i] == 0) {
        c /= facs[n][i];
        x++;
      }
      sol = (sol * 1LL * dp[r][x]) % mod;
    }
    printf("%d\n", sol);
  }
}