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 MAX_N = 1000005, MOD = 1000000007;
int prime[MAX_N], tot, vis[MAX_N];
vector<int> V[MAX_N];
void sieve() {
  for (int i = 2; i <= 1000000; ++i) {
    if (!vis[i]) prime[++tot] = i;
    for (int j = 1; j <= tot && prime[j] * i <= 1000000; ++j) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= tot; ++i) {
    int p = prime[i];
    for (int j = p; j <= 1000000; j += p) {
      int tmp = j, cnt = 0;
      while (tmp % p == 0) tmp /= p, cnt++;
      V[j].push_back(cnt);
    }
  }
}
int f[1000001][20];
int main() {
  sieve();
  int Q;
  scanf("%d", &Q);
  f[0][0] = 1;
  for (int i = 1; i < 20; ++i) f[0][i] = 2;
  for (int i = 1; i <= 1000000; ++i) {
    int sum = f[i - 1][0];
    for (int j = 1; j <= 20; ++j) {
      sum = (sum + f[i - 1][j]) % MOD;
      f[i][j] = sum;
    }
    f[i][0] = 1;
  }
  while (Q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int res = 1;
    for (int i = 0; i < V[n].size(); ++i) res = 1LL * res * f[r][V[n][i]] % MOD;
    printf("%d\n", res);
  }
  return 0;
}