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 mod = (int)1e9 + 7;
const int N = (int)1e6 + 6;
int f[30][N];
int pref[30][N];
int lp[N];
void precalc() {
  pref[0][0] = f[0][0] = 1;
  for (int i = 1; i < 30; i++) {
    f[i][0] = 2;
    pref[i][0] = f[i][0] + pref[i - 1][0];
    pref[i][0] -= pref[i][0] >= mod ? mod : 0;
  }
  for (int level = 1; level < N; level++) {
    pref[0][level] = f[0][level] = 1;
    for (int pw = 1; pw < 30; pw++) {
      f[pw][level] = pref[pw][level - 1];
      pref[pw][level] = f[pw][level] + pref[pw - 1][level];
      pref[pw][level] -= pref[pw][level] >= mod ? mod : 0;
    }
  }
  for (int i = 2; i < N; i++) {
    if (lp[i]) continue;
    for (int j = i; j < N; j += i) {
      lp[j] = lp[j] ? lp[j] : i;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  precalc();
  int T;
  scanf("%d", &T);
  while (T--) {
    int r, n;
    scanf("%d%d", &r, &n);
    if (n == 1) {
      printf("1\n");
      continue;
    }
    long long ans = 1;
    while (n > 1) {
      int p = lp[n];
      int pw = 0;
      while (lp[n] == p) {
        n /= p;
        ++pw;
      }
      ans = ans * f[pw][r] % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}