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 MAXN = 1e6 + 10, MOD = 1e9 + 7;
long long f[MAXN * 2], fr[MAXN + 100], ans;
vector<pair<int, int> > pi;
int q, r, n, lst[MAXN];
long long rev(long long a) {
  int t = MOD - 2;
  long long res = 1;
  for (t; t; t /= 2) {
    if (t % 2) res *= a, res %= MOD;
    a *= a, a %= MOD;
  }
  return res;
}
long long ch(int n, int k) { return f[n] * fr[n - k] % MOD * fr[k] % MOD; }
void getd(int num) {
  pi.clear();
  while (num != 1) {
    int cl = lst[num], c = 0;
    while (num % cl == 0) num /= cl, c++;
    pi.push_back({cl, c});
  }
}
void solve() {
  cin >> r >> n;
  ans = 1;
  getd(n);
  if (r == 0) ans = (1 << pi.size());
  for (int i = 0; i < pi.size() && r > 0; i++) {
    ans *= (ch(pi[i].second - 1 + r, r) * 2 + ch(pi[i].second + r - 1, r - 1)) %
           MOD;
    ans %= MOD;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  f[0] = 1;
  for (int i = 1; i < 100 + MAXN; i++) f[i] = f[i - 1] * i, f[i] %= MOD;
  for (int i = 0; i < 100 + MAXN; i++) fr[i] = rev(f[i]);
  lst[1] = 1;
  for (int j = 2; j < MAXN; j += 2) lst[j] = 2;
  for (int i = 3; i < MAXN; i += 2)
    for (int j = i; j < MAXN; j += 2 * i)
      if (lst[j] == 0) lst[j] = i;
  cin >> q;
  while (q--) solve();
}