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> fact[1000000 + 5];
int dp[1000000 + 5][22];
void precalc() {
  int N = 1000000;
  for (int i = 2; i <= N; ++i) {
    if (fact[i].size() == 0) {
      for (int j = i; j <= N; j += i) fact[j].push_back(i);
    }
  }
  dp[0][0] = 1;
  for (int i = 1; i <= 20; ++i) dp[0][i] = 2;
  for (int r = 1; r <= 1000000; ++r) {
    dp[r][0] = 1;
    for (int q = 1; q <= 20; ++q) {
      dp[r][q] = (dp[r - 1][q] + dp[r][q - 1]) % 1000000007;
    }
  }
}
int powin(int x, int p) {
  int a = 0;
  while (x % p == 0) {
    ++a;
    x /= p;
  }
  return a;
}
int main() {
  precalc();
  int Q;
  cin >> Q;
  while (Q--) {
    int N, R;
    scanf("%d%d", &R, &N);
    signed long long int ans = 1;
    for (int p : fact[N]) {
      ans = ans * dp[R][powin(N, p)] % 1000000007;
    }
    printf("%lld\n", ans);
  }
}