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 + 5, lg = 21;
const long long int md = 1e9 + 7;
int d[MAXN];
long long int dp[MAXN][lg];
bool pr[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  pr[1] = pr[0] = true;
  for (int i = 2; i < MAXN; i++) {
    if (pr[i]) continue;
    d[i] = i;
    for (int j = 2 * i; j < MAXN; j += i) pr[j] = true, d[j] = i;
  }
  for (int i = 0; i < MAXN; i++) dp[i][0] = 1ll;
  for (int i = 1; i < lg; i++) dp[0][i] = 2ll;
  for (int i = 1; i < MAXN; i++)
    for (int j = 1; j < lg; j++) dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % md;
  int q;
  cin >> q;
  while (q--) {
    int n, r;
    cin >> r >> n;
    long long int ans = 1ll;
    while (n > 1) {
      int cnt = 0, n2 = n;
      while (n2 > 1 && (n2 % d[n]) == 0) n2 /= d[n], cnt++;
      n = n2;
      ans = (ans * dp[r][cnt]) % md;
    }
    cout << ans << '\n';
  }
  return 0;
}