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;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
}
const long long MOD = 1e9 + 7;
const int MAX = 1e6 + 7;
const int LN = 21;
long long dp[LN][MAX];
long long solve(int k, int r) {
  if (k == 0)
    return 1;
  else if (r == 0) {
    return 2;
  }
  long long& ret = dp[k][r];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i <= k; i++) {
    ret = (ret + solve(i, r - 1));
    if (ret >= MOD) ret -= MOD;
  }
  return ret;
}
bool v[MAX];
int sp[MAX];
void Sieve() {
  for (int i = 2; i < MAX; i += 2) sp[i] = 2;
  for (long long i = 3; i < MAX; i += 2) {
    if (!v[i]) {
      sp[i] = i;
      for (long long j = i; (j * i) < 1ll * MAX; j += 2) {
        if (!v[j * i]) {
          v[j * i] = true, sp[j * i] = i;
        }
      }
    }
  }
}
int main(int argc, char* argv[]) {
  io();
  Sieve();
  for (int k = 0; k < LN; k++) {
    for (int r = 0; r < MAX; r++) {
      if (k == 0) {
        dp[k][r] = 1;
      } else if (r == 0) {
        dp[k][r] = 2;
      } else {
        dp[k][r] = 0;
        for (int i = 0; i <= k; i++) {
          dp[k][r] = (dp[k][r] + dp[i][r - 1]);
          if (dp[k][r] >= MOD) dp[k][r] -= MOD;
        }
      }
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int n, r;
    cin >> r >> n;
    long long ans = 1;
    while (n != 1) {
      int spr = sp[n];
      int k = 0;
      while (n % spr == 0) {
        k++;
        n /= spr;
      }
      ans = (ans * dp[k][r]) % MOD;
    }
    cout << ans << '\n';
  }
  return 0;
}