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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 1e6 + 100;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
long long dp[maxn][25];
vector<int> vec[maxn];
int n, r;
int q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i < maxn; i++) {
    if (vec[i].empty()) {
      for (int j = i; j < maxn; j += i) {
        vec[j].push_back(i);
        ;
      }
    }
  }
  for (int i = 0; i < maxn; i++) {
    dp[i][0] = 1;
    if (i < 25) {
      dp[0][i] = 2;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i < 24; i++) {
    for (int j = 1; j < maxn; j++) {
      dp[j][i] = (dp[j - 1][i] + dp[j][i - 1]) % mod;
    }
  }
  cin >> q;
  while (q--) {
    cin >> r >> n;
    long long ans = 1;
    for (int i : vec[n]) {
      int cnt = 0;
      while (n % i == 0) {
        cnt++;
        n /= i;
      }
      ans *= dp[r][cnt];
      ans %= mod;
    }
    cout << ans << '\n';
  }
  return (0);
}