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:

<image>

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;
namespace Solution {
int p[((long long)1e6 + 10)] = {0};
void sieve() {
  for (long long i = 2; i < ((long long)1e6 + 10); ++i)
    if (p[i] == 0) {
      p[i] = i;
      for (long long f = i * i; f < ((long long)1e6 + 10); f += i) p[f] = i;
    }
}
vector<vector<int> > dp;
void preprocess() {
  dp.resize(((long long)1e6 + 10), vector<int>((20), 2));
  dp[0][0] = 1;
  for (int i = 1; i < ((long long)1e6 + 10); ++i) {
    dp[i][0] = 1;
    for (int f = 1; f < (20); ++f)
      dp[i][f] = ((long long)dp[i - 1][f] + (long long)dp[i][f - 1]) %
                 ((long long)1e9 + 7);
  }
}
long long f(int r, int n) {
  map<int, int> fcnt;
  long long ans = 1;
  for (; n > 1; n /= p[n]) fcnt[p[n]]++;
  for (pair<int, int> i : fcnt) {
    ans = (ans * dp[r][i.second]) % ((long long)1e9 + 7);
  }
  return ans;
}
int main() {
  sieve();
  preprocess();
  int t;
  cin >> t;
  while (t--) {
    int n, r;
    cin >> r >> n;
    cout << f(r, n) << '\n';
  }
  return 0;
}
}  // namespace Solution
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  return Solution::main();
}