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 add(int a, int b) { return a + b - (a + b >= MOD) * MOD; }
int sub(int a, int b) { return a - b + (a - b < 0) * MOD; }
int mult(int a, int b) { return (1LL * a * b) % MOD; }
int binpow(int a, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = mult(ans, a);
    p >>= 1;
    a = mult(a, a);
  }
  return ans;
}
int inv(int x) { return binpow(x, MOD - 2); }
int fact[2000013], ifact[2000013];
int low[1000013];
vector<int> has[1000013];
int combo(int x, int y) {
  if (x < y) return 0;
  return mult(fact[x], mult(ifact[y], ifact[x - y]));
}
void precalc() {
  fact[0] = ifact[0] = 1;
  for (int i = 1; i < 2000013; i++) {
    fact[i] = mult(fact[i - 1], i);
    ifact[i] = inv(fact[i]);
  }
  iota(low, low + 1000013, 0);
  for (int i = 2; i < 1000013; i++) {
    if (low[i] == i) {
      has[i] = {1};
      for (int j = i + i; j < 1000013; j += i) low[j] = min(low[j], i);
    } else {
      int j = i;
      int num = 0;
      while (j % low[i] == 0) {
        j /= low[i];
        num += 1;
      }
      has[i] = has[j];
      has[i].push_back(num);
    }
  }
}
int q, r, n;
int main() {
  precalc();
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    if (r == 0)
      printf("%d\n", binpow(2, has[n].size()));
    else {
      int ans = 1;
      for (int i : has[n]) {
        int cur = add(combo(r + i - 1, i - 1), combo(r + i, i));
        ans = mult(ans, cur);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}