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:

<image>

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>
inline int mul(int a, int b, int md) { return (a * 1ll * b) % md; }
int main() {
  int q;
  scanf("%d", &q);
  const int md = 1000000007;
  const int mv = 1000005;
  std::vector<std::vector<int> > dp(mv, std::vector<int>(25, 0));
  for (int i = 0; i < dp.size(); i++) dp[i][0] = 1;
  for (int j = 1; j < dp[0].size(); j++) dp[0][j] = 2;
  for (int i = 1; i < dp.size(); i++) {
    for (int j = 1; j < dp[0].size(); j++) {
      dp[i][j] += dp[i - 1][j];
      dp[i][j] += dp[i][j - 1];
      if (dp[i][j] >= md) dp[i][j] -= md;
    }
  }
  std::vector<int> lowest(mv);
  for (int i = 2; i < lowest.size(); i++) {
    if (lowest[i] == 0) {
      for (int j = i; j < lowest.size(); j += i) {
        if (lowest[j] == 0) {
          lowest[j] = i;
        }
      }
    }
  }
  for (int it = 0; it < q; it++) {
    int r, n;
    scanf("%d %d", &r, &n);
    int ans = 1;
    while (n > 1) {
      int p = lowest[n];
      int cnt = 0;
      while (n % p == 0) {
        n /= p;
        cnt++;
      }
      ans = mul(ans, dp[r][cnt], md);
    }
    printf("%d\n", ans);
  }
}