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:

<image>

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, base = 1e9 + 7;
int test, n, r, sieve[maxN];
long long val[21][maxN];
pair<int, int> f[maxN];
vector<int> prime;
void eratos() {
  for (int i = 2; i * i < maxN; i++)
    if (sieve[i] == 0)
      for (int j = i * i; j < maxN; j += i) sieve[j] = i;
  for (int i = 2; i < maxN; i++)
    if (sieve[i] == 0) prime.push_back(i);
}
void init() {
  eratos();
  for (int i = 2; i < maxN; i++)
    if (sieve[i] == 0)
      f[i] = pair<int, int>(i, 1);
    else {
      int p = sieve[i];
      if (f[i / p].first == p)
        f[i] = pair<int, int>(p, f[i / p].second + 1);
      else
        f[i] = pair<int, int>(-1, 0);
    }
  for (int i = 1; i <= 20; i++) val[i][0] = 2;
  for (int i = 1; i <= 20; i++)
    for (int j = 1; j < maxN; j++) {
      if (i == 1)
        val[i][j] = val[i][j - 1] + 1;
      else
        val[i][j] = (val[i - 1][j] + val[i][j - 1]) % base;
    }
}
int main() {
  init();
  scanf("%d", &test);
  while (test--) {
    scanf("%d%d", &r, &n);
    long long res = 1;
    for (int i = 0; i < prime.size() && n > 1; i++) {
      if (n % prime[i] == 0) {
        int v = 1;
        while (n % prime[i] == 0) n /= prime[i], v *= prime[i];
        res = (res * val[f[v].second][r]) % base;
      }
      if (f[n].second) {
        res = (res * val[f[n].second][r]) % base;
        break;
      }
    }
    cout << res << "\n";
  }
}