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;
string itos(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
int dp[1000010][25], q;
vector<int> p;
vector<int> f[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  scanf("%d", &q);
  for (int i = 0; i < 1000010; i++) {
    for (int j = 0; j < 20; j++) {
      if (j == 0) {
        dp[i][j] = 1;
        continue;
      }
      if (i == 0) {
        dp[i][j] = 2;
        continue;
      }
      (dp[i][j] += dp[i - 1][j] + dp[i][j - 1]) %= 1000000007;
    }
  }
  for (int i = 2; i < 1000010; i++) {
    if (f[i].empty()) {
      p.push_back(i);
      for (int j = i; j < 1000010; j += i) f[j].push_back(i);
    }
  }
  while (q--) {
    int r, n;
    long long res = 1;
    scanf("%d%d", &r, &n);
    for (auto x : f[n]) {
      int now = 0;
      while (n % x == 0) {
        now++;
        n /= x;
      }
      (res *= dp[r][now]) %= 1000000007;
    }
    printf("%I64d\n", res);
  }
  return 0;
}