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 = 1000005;
const int MOD = 1000000007;
int pr[MAXN], dp[MAXN][20];
inline long long fun(int a, int r) {
  long long ans = 1;
  while (a != 1) {
    int c = 0, cur = pr[a];
    while (pr[a] == cur) a /= pr[a], c++;
    ans = ans * dp[r][c] % MOD;
  }
  return ans;
}
int main() {
  int q;
  scanf("%d", &q);
  for (int i = 2; i < MAXN; i++) {
    if (pr[i] == 0) {
      for (int j = i; j < MAXN; j += i)
        if (pr[j] == 0) pr[j] = i;
    }
  }
  dp[0][0] = 1;
  for (int i = 1; i < 20; i++) dp[0][i] = 2;
  for (int i = 1; i < MAXN; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < 20; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
      if (dp[i][j] >= MOD) dp[i][j] -= MOD;
    }
  }
  while (q--) {
    int r, n;
    scanf("%d %d", &r, &n);
    long long ans = fun(n, r);
    cout << ans << '\n';
  }
  return 0;
}