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;
int q;
long long r, n;
vector<int> degp[1000005];
long long answ[1000005][21];
long long m = 1000000007;
long long powmod(long long x, long long po) {
  if (po == 0) {
    return 1;
  }
  long long res = powmod(x, po / 2);
  res = (res * res) % m;
  if (po % 2 == 1) {
    res = (res * x) % m;
  }
  return res;
}
int main() {
  for (int i = 2; i <= 1000000; i++) {
    if (degp[i].empty()) {
      long long t = i;
      long long u = 1;
      while (t <= 1000000) {
        for (int j = 1; j <= 1000000 / t; j++) {
          if (j % i != 0) {
            degp[t * j].push_back(u);
          }
        }
        u++;
        t = t * (long long)i;
      }
    }
  }
  for (int i = 0; i <= 1000000; i++) {
    answ[i][0] = 1;
  }
  for (int i = 1; i < 21; i++) {
    answ[0][i] = 2;
    for (int j = 1; j <= 1000000; j++) {
      answ[j][i] = (answ[j - 1][i] + answ[j][i - 1]) % m;
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%I64d%I64d", &r, &n);
    long long res = 1;
    for (int k = 0; k < degp[n].size(); k++) {
      res = (res * answ[r][degp[n][k]]) % m;
    }
    printf("%I64d\n", res);
  }
}