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 dp[1000005][25];
int p[1000005];
int sum(long long a, long long b) {
  long long ans = (a + b);
  if (ans > 1000000007) ans %= 1000000007;
  return int(ans);
}
int mul(long long a, long long b) {
  long long ans = (a * b);
  if (ans > 1000000007) ans %= 1000000007;
  return int(ans);
}
void sieve() {
  for (int i = 0; i < 1000005; i++) p[i] = i;
  for (int i = 2; i < 1000005; i++) {
    if (p[i] == i) {
      for (int j = i + i; j < 1000005; j += i) p[j] = i;
    }
  }
}
int main() {
  int n, r, ans = 0, tc;
  sieve();
  for (int i = 0; i < 1000005; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < 25; j++) {
      if (i == 0)
        dp[0][j] = 2;
      else
        dp[i][j] = sum(dp[i][j - 1], dp[i - 1][j]);
    }
  }
  scanf("%d", &tc);
  while (tc--) {
    ans = 1;
    scanf("%d%d", &r, &n);
    while (n != 1) {
      int cnt = 0;
      int k = p[n];
      while (n % k == 0) {
        n /= k;
        cnt++;
      }
      ans = mul(ans, dp[r][cnt]);
    }
    printf("%d\n", ans);
  }
}