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 N = 1e6 + 10, P = 1e9 + 7;
long long dp[21][N], lp[N];
long long n;
void build1() {
  lp[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!lp[i]) {
      for (int j = i; j < N; j += i) {
        lp[j] = i;
      }
    }
  }
}
void build2() {
  for (int i = 0; i < N; i++) dp[0][i] = 1;
  for (int i = 0; i <= 20; i++) dp[i][0] = 2;
  for (int i = 1; i <= 20; i++) {
    for (int j = 1; j < N; j++) {
      dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % P;
    }
  }
}
long long f(long long a, long long b) {
  long long ans = 1;
  while (b > 1) {
    long long cnt = 0, p = lp[b];
    while (b % p == 0) {
      cnt++;
      b /= p;
    }
    ans = (ans * dp[cnt][a]) % P;
  }
  return ans;
}
int main() {
  build1();
  build2();
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", f(a, b));
  }
}