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 int N = 1000005;
const int mod = 1e9 + 7;
int data, r, n, pr[N];
int dp[N][20];
using namespace std;
inline int read() {
  char c;
  int x;
  while (c = getchar(), c < '0' || c > '9')
    ;
  x = c ^ '0';
  while (c = getchar(), c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ '0');
  return x;
}
int query() {
  int ans = 1;
  while (n != 1) {
    int p = pr[n], k = 0;
    while (n % p == 0) n /= p, k++;
    ans = (long long)ans * dp[r][k] % mod;
  }
  return ans;
}
int main() {
  n = 1000000;
  int i, j;
  for (i = 2; i <= n; i++)
    if (!pr[i])
      for (j = i; j <= n; j += i) pr[j] = i;
  dp[0][0] = 1;
  for (i = 1; i < 20; i++) dp[0][i] = 2;
  for (i = 1; i <= n; i++)
    for (dp[i][0] = dp[i - 1][0], j = 1; j < 20; j++)
      dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
  scanf("%d", &data);
  while (data--) {
    r = read(), n = read();
    printf("%d\n", query());
  }
  return 0;
}