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 long long mod = 1e9 + 7;
int pri[1000000 + 5];
int b[1000000 + 5];
void getPrime() {
  for (int i = 2; i <= 1000000; i++) {
    if (pri[i] == 0) {
      pri[++pri[0]] = i;
      b[i] = i;
    }
    for (int j = 1; j <= pri[0] && pri[j] <= 1000000 / i; j++) {
      pri[pri[j] * i] = 1;
      b[pri[j] * i] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
int dp[1000005][21];
int q, r, n;
int main() {
  dp[0][0] = 1;
  for (int j = 1; j <= 20; j++) {
    dp[0][j] = 2;
  }
  for (int i = 1; i <= 1000000; i++) {
    int tmp = 0;
    for (int j = 0; j <= 20; j++) {
      tmp = (tmp + dp[i - 1][j]) % mod;
      dp[i][j] = tmp;
    }
  }
  getPrime();
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    long long tmp = 1;
    int x = n;
    while (x > 1) {
      int p = b[x];
      int cnt = 0;
      while (b[x] == p) {
        x /= p;
        cnt++;
      }
      tmp = tmp * dp[r][cnt] % mod;
    }
    printf("%I64d\n", tmp);
  }
  return 0;
}