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;
const int FMAX = 1100000, PMAX = 22, MOD = 1000000007;
int Q, N, R, ans;
int coe[FMAX][PMAX];
int factor[FMAX];
vector<int> prime;
inline int quickModPow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (long long)res * x % MOD;
    x = (long long)x * x % MOD;
    y >>= 1;
  }
  return res;
}
inline int C(int n, int m) {
  if (m < 0 || n < m) return 0;
  return coe[n][m];
}
void genPrime() {
  int i, j;
  for (i = 2; i < FMAX; i += 1) {
    if (!factor[i]) {
      factor[i] = i;
      prime.push_back(i);
    }
    for (j = 0; j < (int)prime.size() && i * prime[j] < FMAX; j += 1) {
      factor[i * prime[j]] = prime[j];
      if (factor[i] == prime[j]) break;
    }
  }
}
int main() {
  int i, j, f, t;
  coe[0][0] = 1;
  for (i = 1; i < FMAX; i += 1) {
    coe[i][0] = 1;
    for (j = 1; j < PMAX; j += 1)
      coe[i][j] = (coe[i - 1][j - 1] + coe[i - 1][j]) % MOD;
  }
  genPrime();
  scanf("%d", &Q);
  for (i = 1; i <= Q; i += 1) {
    ans = 1;
    scanf("%d %d", &R, &N);
    while (N > 1) {
      t = factor[N];
      for (f = 0; !(N % t); f += 1, N /= t)
        ;
      ans = ans * ((long long)C(f + R - 1, f) + 2 * C(f + R - 1, f - 1)) % MOD;
    }
    printf("%d\n", ans);
  }
  exit(0);
}