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 size = 1000 * 1000 + 100;
const int ssize = 21;
const long long mdl = 1000 * 1000 * 1000 + 7;
int q;
long long help[ssize][size];
int mindiv[size];
int nod(int a, int b) {
  if (b == 0) {
    return a;
  }
  return nod(b, a % b);
}
int main() {
  for (int i = 0; i < size; i++) mindiv[i] = i;
  for (int i = 2; i * i < size; i++)
    if (mindiv[i] == i) {
      for (int j = i; i * j < size; j++) {
        mindiv[i * j] = min(mindiv[i * j], i);
      }
    }
  for (int i = 0; i < size; i++) help[0][i] = 1;
  for (int i = 1; i < ssize; i++) {
    help[i][0] = 2;
    for (int j = 1; j < size; j++) {
      help[i][j] = help[i - 1][j] + help[i][j - 1];
      if (help[i][j] >= mdl) {
        help[i][j] -= mdl;
      }
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int r, n;
    scanf("%d%d", &r, &n);
    int lst = -1;
    int cnt = 0;
    long long ans = 1;
    while (n > 1) {
      if (mindiv[n] != lst) {
        ans = (ans * help[cnt][r]) % mdl;
        lst = mindiv[n];
        cnt = 1;
      } else {
        cnt++;
      }
      n /= mindiv[n];
    }
    ans = (ans * help[cnt][r]) % mdl;
    printf("%lld\n", ans);
  }
  return 0;
}