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 MAX = 2e6 + 7, P = 1e9 + 7;
int Px[MAX], f[MAX], iv[MAX];
vector<int> pr;
int fpow(int b, int p) {
  long long int r = 1;
  while (p) {
    if (p & 1) r *= b, r %= P;
    b = 1ll * b * b % P;
    p >>= 1;
  }
  return r;
}
void pre() {
  for (int i = 2; i < MAX; ++i) {
    if (!Px[i]) {
      Px[i] = i;
      pr.emplace_back(i);
    }
    for (int p : pr) {
      if (1ll * p * i >= MAX) break;
      Px[p * i] = p;
      if (i % p == 0) break;
    }
  }
  f[0] = 1, iv[0] = 1;
  for (int i = 1; i < MAX; ++i) {
    f[i] = 1ll * f[i - 1] * i % P;
    iv[i] = fpow(f[i], P - 2);
  }
}
int C(int n, int k) {
  int r = 1ll * f[n] * iv[k] % P;
  return 1ll * r * iv[n - k] % P;
}
int calc(int e, int r) {
  int res = 0;
  for (int i = 0; i <= e; ++i) {
    res += (i ? 2ll * C(e + r - 1 - i, r - 1) % P : C(e + r - 1 - i, r - 1));
    if (res >= P) res -= P;
  }
  return res;
}
int fun(int n, int r) {
  int ans = 1;
  while (Px[n]) {
    int cp = Px[n], e = 0;
    while (n % cp == 0) {
      n /= cp;
      ++e;
    }
    if (r == 0)
      ans = 2ll * ans % P;
    else
      ans = 1ll * ans * calc(e, r) % P;
  }
  return ans;
}
int main() {
  pre();
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    printf("%d\n", fun(n, r));
  }
  return 0;
}