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;
long long low[1000005], dp[1000005][20], sum[20] = {1};
int main() {
  long long i, j, r, n, q, ans, cnt, p;
  for (i = 2; i <= 1000000; i++)
    if (!low[i])
      for (j = i; j <= 1000000; j += i) low[j] = i;
  for (i = 0; i <= 1000000; i++) dp[i][0] = 1;
  for (i = 1; i < 20; i++) {
    dp[0][i] = 2;
    sum[i] = sum[i - 1] + dp[0][i];
  }
  for (i = 1; i <= 1000000; i++) {
    for (j = 1; j < 20; j++) {
      dp[i][j] = sum[j];
      sum[j] = (sum[j - 1] + dp[i][j]) % 1000000007;
    }
  }
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld%lld", &r, &n);
    ans = 1;
    while (n != 1) {
      cnt = 0, p = low[n];
      while (n % p == 0) {
        cnt++;
        n /= p;
      }
      ans = ans * dp[r][cnt] % 1000000007;
    }
    cout << ans << endl;
  }
  return 0;
}