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 LG = 23, N = 1e6 + 20, mod = 1e9 + 7;
int q, f[N][LG], res[N], d[N];
inline int sum(int a, int b) {
  int res = a + b;
  if (res >= mod) res -= mod;
  return res;
}
void pre() {
  f[0][0] = 1;
  for (int i = 1; i < LG; i++) f[0][i] = 2;
  for (int i = 1; i < N; i++) {
    f[i][0] = f[i - 1][0];
    for (int j = 1; j < LG; j++) f[i][j] = sum(f[i - 1][j], f[i][j - 1]);
  }
  memset(d, -1, sizeof d);
  for (int i = 2; i < N; i++)
    for (int j = i + i; j < N; j += i)
      if (d[j] == -1) d[j] = i;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  pre();
  cin >> q;
  while (q--) {
    int r, n, res = 1;
    cin >> r >> n;
    while (n > 1) {
      int k = d[n], cnt = 0;
      if (k == -1)
        n = 1, cnt = 1;
      else
        while (n % k == 0) n /= k, cnt++;
      res = res * 1ll * f[r][cnt] % mod;
    }
    cout << res << '\n';
  }
}