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>
const int64_t LIM = 1000050;
const int64_t MOD = 1e9 + 7;
std::vector<int64_t> exps[LIM + 1];
int64_t factorial[LIM + 1];
int64_t inverse[LIM + 1];
int64_t inversefactorial[LIM + 1];
int64_t combo(int64_t a, int64_t b) {
  return ((a < 0) || (b < 0)) ? 0
                              : (factorial[a + b] * inversefactorial[a] % MOD *
                                 inversefactorial[b] % MOD);
}
int main() {
  for (int64_t i = 2; i <= LIM; i++) {
    if (exps[i].empty()) {
      for (int64_t k = i; k <= LIM; k += i) {
        exps[k].push_back(1);
      }
      for (int64_t p = i * i; p <= LIM; p *= i) {
        for (int64_t k = p; k <= LIM; k += p) {
          exps[k].back()++;
        }
      }
    }
  }
  inverse[1] = 1;
  for (int64_t i = 2; i <= LIM; i++) {
    inverse[i] = (MOD - MOD / i) * inverse[MOD % i] % MOD;
  }
  factorial[0] = 1;
  inversefactorial[0] = 1;
  for (int64_t i = 1; i <= LIM; i++) {
    factorial[i] = factorial[i - 1] * i % MOD;
    inversefactorial[i] = inversefactorial[i - 1] * inverse[i] % MOD;
  }
  int Q;
  scanf("%d", &Q);
  for (int64_t i = 0; i < Q; i++) {
    int R, N;
    scanf("%d %d", &R, &N);
    int64_t prod = 1;
    for (int64_t e : exps[N]) {
      prod = prod * (2 * combo(e, R) - combo(e, R - 1) + MOD) % MOD;
    }
    printf("%d\n", static_cast<int>(prod));
  }
  return 0;
}