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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 2;
const int maxp = 20 + 2;
const int R = 1e6;
const int P = 20;
long long dp[maxn][maxp];
long long lowest_prime[maxn];
long long solve(long long r, long long n) {
  long long res = 1;
  while (n > 1) {
    long long p = lowest_prime[n];
    long long cnt = 0;
    while (n % p == 0) {
      n = n / p;
      cnt++;
    }
    res = res * dp[r][cnt] % mod;
  }
  res %= mod;
  res += mod;
  res %= mod;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 2; i <= R; i++) {
    if (!lowest_prime[i]) {
      for (int j = i; j <= R; j += i) {
        lowest_prime[j] = i;
      }
    }
  }
  dp[0][0] = 1;
  for (int j = 1; j <= P; j++) {
    dp[0][j] = 2;
  }
  for (int r = 1; r <= R; r++) {
    for (int i = 0; i <= P; i++) {
      dp[r][i] = (dp[r - 1][i] + (i > 0 ? dp[r][i - 1] : 0)) % mod;
    }
  }
  int t;
  cin >> t;
  while (t--) {
    long long r, n;
    cin >> r >> n;
    cout << solve(r, n) << "\n";
  }
  return 0;
}