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 mod = 1e9 + 7;
int dp[23][1000001];
int pref[23][1000001];
int sieve[1001000];
int main(void) {
  for (int i = 1; i <= 20; i++) {
    dp[i][0] = 2;
    pref[i][0] = (dp[i][0] + pref[i - 1][0]) % mod;
  }
  for (int r = 1; r <= 1e6; r++) {
    for (int i = 1; i <= 20; i++) {
      dp[i][r] = (pref[i][r - 1] + 1) % mod;
      pref[i][r] = (dp[i][r] + pref[i - 1][r]) % mod;
    }
  }
  sieve[1] = 1;
  for (int i = 2; i <= 1e6; i++) {
    if (sieve[i] == 0) {
      for (int j = i; j <= 1e6; j += i) sieve[j] = sieve[j] == 0 ? i : sieve[j];
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int r, n;
    scanf("%d %d", &r, &n);
    if (n == 1) {
      printf("1\n");
      continue;
    }
    vector<int> p, mul;
    int x = n;
    while (x != 1) {
      if (p.empty() || p.back() != sieve[x]) {
        p.push_back(sieve[x]);
        mul.push_back(1);
      } else {
        mul[mul.size() - 1]++;
      }
      x /= sieve[x];
    }
    int ans = 1;
    for (int i = 0; i < p.size(); i++)
      ans = (ans * (long long)dp[mul[i]][r]) % mod;
    printf("%d\n", ans);
  }
}