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 fact[2000001];
long long inv[2000001];
vector<int> pr = {2};
long long nCr(int n, int r) {
  long long val = fact[n];
  val = val * inv[r] % 1000000007;
  val = val * inv[n - r] % 1000000007;
  return val;
}
int main() {
  fact[0] = 1;
  inv[0] = 1;
  for (int i = 1; i <= 2000000; i++) {
    fact[i] = fact[i - 1] * i % 1000000007;
  }
  inv[1] = 1;
  for (int i = 2; i <= 2000000; i++) {
    int left = 1000000007 - 2;
    long long res = 1;
    long long gop = i;
    while (left) {
      if (left & 1) {
        res *= gop;
        res %= 1000000007;
      }
      gop *= gop;
      gop %= 1000000007;
      left >>= 1;
    }
    inv[i] = res * inv[i - 1] % 1000000007;
  }
  for (int i = 3; i <= 2000; i += 2) {
    bool prime = true;
    for (int j = 0; pr[j] * pr[j] <= i; j++) {
      if (i % pr[j] == 0) {
        prime = false;
        break;
      }
    }
    if (prime) pr.push_back(i);
  }
  int t, n, r;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &r, &n);
    long long ans = 1;
    for (int j = 0; pr[j] * pr[j] <= n; j++) {
      int i = pr[j];
      int cnt = 0;
      while (n % i == 0) {
        cnt++;
        n /= i;
      }
      if (cnt) {
        if (r == 0)
          ans *= 2;
        else {
          long long cal =
              (2LL * nCr(r + cnt - 1, cnt - 1) + nCr(r + cnt - 1, cnt)) %
              1000000007;
          ans = ans * cal % 1000000007;
        }
      }
    }
    if (n > 1) {
      ans = ans * (r + 2) % 1000000007;
    }
    printf("%lld\n", ans);
  }
}