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;
const int MOD = (int)1e9 + 7;
int sum(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int mult(int a, int b) { return (a * 1LL * b) % MOD; }
int bin_pow(int a, int n) {
  int res = 1;
  for (; n; n >>= 1, a = mult(a, a))
    if (n & 1) res = mult(res, a);
  return res;
}
int inv(int a) { return bin_pow(a, MOD - 2); }
const int C = (int)1e6 + 100;
namespace Combi {
int f[C], rf[C];
void init(int n) {
  f[0] = 1;
  for (int i = 1; i <= n; ++i) f[i] = mult(f[i - 1], i);
  rf[n] = inv(f[n]);
  for (int i = n; i > 0; --i) rf[i - 1] = mult(rf[i], i);
}
int C(int n, int k) { return mult(f[n], mult(rf[k], rf[n - k])); }
}  // namespace Combi
int divide(int& n, int d) {
  int res = 0;
  while (n % d == 0) {
    n /= d;
    ++res;
  }
  return res;
}
int p[C];
int main() {
  iota(p, p + C, 0);
  for (int x = 2; x < C; ++x) {
    if (p[x] != x) continue;
    for (int y = x + x; y < C; y += x) p[y] = x;
  }
  Combi::init(C);
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d %d", &r, &n);
    int ans = 1;
    if (r == 0) {
      while (n != 1) {
        divide(n, p[n]);
        ans = sum(ans, ans);
      }
    } else {
      int invR = inv(r);
      while (n != 1) {
        int alpha = divide(n, p[n]);
        int fact = sum(mult(2 * alpha, invR), 1);
        fact = mult(fact, Combi::C(alpha + r - 1, alpha));
        ans = mult(ans, fact);
      }
    }
    printf("%d\n", ans);
  }
}