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 MAX = 1e6 + 1, MOD = 1e9 + 7;
int n, r, t, f[MAX][20], p[MAX];
int main() {
  f[0][0] = 1;
  for (int i = 1; i < 20; i++) f[0][i] = 2;
  for (int i = 1; i < MAX; i++) {
    f[i][0] = 1;
    for (int j = 1; j < 20; j++) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % MOD;
  }
  for (int i = 4; i < MAX; i += 2) p[i] = 2;
  for (int i = 3; i * i < MAX; i += 2) {
    if (p[i] > 0) continue;
    for (int j = 2 * i; j < MAX; j += i) p[j] = i;
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &r, &n);
    vector<int> v;
    while (p[n]) {
      v.push_back(p[n]);
      n /= p[n];
    }
    if (n > 1) v.push_back(n);
    sort(v.begin(), v.end());
    int c = 1, sol = 1;
    for (int i = 0; i < (int)v.size(); i++) {
      if (i == (int)v.size() - 1 || v[i] != v[i + 1]) {
        sol = (sol * 1LL * f[r][c]) % MOD;
        c = 1;
      } else
        c++;
    }
    printf("%d\n", sol);
  }
  return 0;
}