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;
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
void read(vector<T> &v) {
  T x;
  cin >> x;
  v.push_back(x);
}
int f[1000005][20], lp[1000005];
int main() {
  for (int i = (2); i <= (1000005 - 1); ++i) {
    if (lp[i] == 0) {
      for (int j = (i); j <= (1000005 - 1); j += i) {
        if (!lp[j]) lp[j] = i;
      }
    }
  }
  for (int j = (0); j <= (20 - 1); ++j) f[0][j] = (j == 0) ? 1 : 2;
  for (int i = (1); i <= (1000005 - 1); ++i) {
    f[i][0] = f[i - 1][0];
    for (int j = (1); j <= (20 - 1); ++j)
      f[i][j] = (f[i][j - 1] + f[i - 1][j]) % 1000000007;
  }
  int q;
  scanf("%d", &q);
  for (int i = (1); i <= (q); ++i) {
    int r, n;
    scanf("%d%d", &r, &n);
    int ans = 1;
    while (lp[n]) {
      int x = lp[n], cnt = 0;
      while (lp[n] == x) ++cnt, n /= x;
      ans = ((long long)ans * f[r][cnt]) % 1000000007;
    }
    printf("%d\n", ans);
  }
  return 0;
}