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 MX = 1200000;
const long long MOD = 1e9 + 7;
long long fc[MX];
long long bfc[MX];
vector<int> pr;
int bb[MX];
long long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
    --b;
    ans = (ans * a) % MOD;
  }
  return ans;
}
long long cnk(int n, int k) {
  if (k > n) return 0;
  return fc[n] * bfc[k] % MOD * bfc[n - k] % MOD;
}
int main() {
  for (int i = 0; i < MX; ++i) bb[i] = i;
  for (int i = 2; i < MX; ++i) {
    if (bb[i] == i) pr.push_back(i);
    for (int j : pr) {
      if (j * i >= MX || j > bb[i]) break;
      bb[j * i] = j;
    }
  }
  fc[0] = 1;
  for (int i = 1; i < MX; ++i) fc[i] = (fc[i - 1] * i) % MOD;
  bfc[MX - 1] = pw(fc[MX - 1], MOD - 2);
  for (int i = MX - 2; i >= 0; --i) bfc[i] = (bfc[i + 1] * (i + 1)) % MOD;
  int q;
  scanf("%d", &q);
  for (int it = 0; it < q; ++it) {
    int r, n;
    scanf("%d%d", &r, &n);
    long long ans = 1;
    while (n != 1) {
      int k = bb[n];
      int cnt = 0;
      while (n % k == 0) n /= k, ++cnt;
      long long dsh = cnk(r + cnt - 1, cnt);
      long long ndsh = 0;
      for (int i = 0; i < cnt; ++i) ndsh = (ndsh + cnk(r + i - 1, i)) % MOD;
      if (r == 0) dsh = 0, ndsh = 1;
      ans = (ans * (dsh + 2 * ndsh)) % MOD;
    }
    printf("%lld\n", ans);
  }
  return 0;
}