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 dp[1000001][21], previ[21];
int prime[1000001];
void factseive() {
  int i, j;
  for (i = 1; i <= 1000000; i++) {
    prime[i] = i;
  }
  for (i = 2; i * i <= 1000000; i++) {
    if (prime[i] == i) {
      for (j = 2 * i; j <= 1000000; j += i) {
        if (prime[j] == j) {
          prime[j] = i;
        }
      }
    }
  }
}
long long pfac(int n, int r) {
  if (prime[n] == n) {
    return dp[r][1];
  }
  long long ans = 1;
  while (n != 1) {
    int cnt = 0, p = prime[n];
    while (n % p == 0) n /= p, cnt++;
    ans = (ans * dp[r][cnt]) % 1000000007;
  }
  return ans;
}
int main() {
  long long int i, j, k, l, n, m, val, ans = 1, r, q;
  dp[0][0] = 1;
  previ[0] = 1;
  for (i = 1; i <= 20; i++) {
    dp[0][i] = 2;
    previ[i] = (previ[i - 1] + 2) % 1000000007;
  }
  for (r = 1; r <= 1000000; r++) {
    for (j = 0; j <= 20; j++) {
      dp[r][j] = previ[j];
      if (j > 0)
        previ[j] = (dp[r][j] + previ[j - 1]) % 1000000007;
      else
        previ[j] = dp[r][j];
    }
  }
  factseive();
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld", &r);
    scanf("%lld", &n);
    if (n == 1) {
      printf("%d\n", 1);
      continue;
    }
    printf("%lld\n", pfac(n, r));
  }
  return 0;
}