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;
vector<int> minimum_prime_fuctors(int x) {
  vector<int> res(x + 1);
  vector<int> primes;
  for (int i = 2; i <= x; i++) {
    if (res[i] == 0) {
      res[i] = i;
      primes.push_back(i);
    }
    for (int j = 0;
         j < primes.size() && primes[j] <= res[i] && primes[j] * i <= x; j++) {
      res[i * primes[j]] = primes[j];
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  vector<int> mp = minimum_prime_fuctors(1010101);
  const int mod = 1000000007;
  int fs = 1010101;
  vector<long long> fact(fs), ifact(fs), invs(fs);
  fact[0] = ifact[0] = fact[1] = ifact[1] = invs[1] = 1;
  for (int i = 2; i < fs; i++) {
    fact[i] = fact[i - 1] * i % mod;
    invs[i] = mod - invs[mod % i] * (mod / i) % mod;
    ifact[i] = ifact[i - 1] * invs[i] % mod;
  }
  function<long long(int, int)> binom = [&](int a, int b) {
    if (b == 0) return 1LL;
    if (a < b || b < 0) return 0LL;
    return fact[a] * ifact[b] % mod * ifact[a - b] % mod;
  };
  while (q--) {
    int r, n;
    cin >> r >> n;
    long long res = 1;
    while (n > 1) {
      int p = mp[n];
      int k = 0;
      while (n % p == 0) {
        n /= p;
        k++;
      }
      long long te = 0;
      for (int i = 0; i <= k; i++) {
        te = (te + binom(r + k - 1 - i, k - i) * (i == 0 ? 1 : 2) % mod) % mod;
      }
      res = res * te % mod;
    }
    cout << res << "\n";
  }
  return 0;
}