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 maxn = 1000000 + 5;
const int mod = 1000000007;
int dp[maxn][21];
int lp[maxn];
void init() {
  lp[1] = 1;
  for (long long i = 2; i < maxn; i++) {
    if (!lp[i]) {
      for (long long j = i; j < maxn; j = j + i) lp[j] = i;
    }
  }
  for (int i = 0; i < 21; i++) dp[0][i] = 2;
  for (int i = 0; i < maxn; i++) dp[i][0] = 1;
  for (int i = 1; i < maxn; i++)
    for (int j = 1; j < 21; j++) dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
}
int main() {
  init();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int r, n;
    scanf("%d%d", &r, &n);
    long long ans = 1;
    int tmp = n;
    int p = lp[tmp];
    while (tmp != 1) {
      int x = 0;
      while (tmp % p == 0) {
        tmp = tmp / p;
        x++;
      }
      ans = (ans * dp[r][x]) % mod;
      p = lp[tmp];
    }
    printf("%d\n", ans);
  }
  return 0;
}