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 long long md = (1e9) + 7, SZ = 1000000;
int t, n, k, divide[SZ + 1] = {};
long long fkt[SZ + 69] = {}, invfkt[SZ + 69] = {};
inline long long pw(long long b, long long e) {
  long long t = 1;
  while (e) {
    if (e & 1) {
      t *= b;
      t %= md;
    }
    b *= b;
    b %= md;
    e >>= 1;
  }
  return t;
}
inline long long c(long long n, long long r) {
  if (n == r) {
    return 1;
  }
  if (r < 0 || n < r) {
    return 0;
  }
  return ((fkt[n] * invfkt[r]) % md * invfkt[n - r]) % md;
}
inline long long getRes(long long cr) {
  long long rs = 1, pv = -1, p = 0;
  while (cr > 1) {
    if (pv != divide[cr]) {
      rs = (rs * ((2 * c(p - 1 + k, k) + c(p + k - 1, k - 1)) % md)) % md;
      pv = divide[cr];
      p = 0;
    }
    p++;
    cr = cr / divide[cr];
  }
  rs = (rs * ((2 * c(p - 1 + k, k) + c(p + k - 1, k - 1)) % md)) % md;
  return rs;
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  fkt[0] = invfkt[0] = 1;
  for (int i = 1; i <= SZ + 60; i++) {
    fkt[i] = (fkt[i - 1] * i) % md;
    invfkt[i] = pw(fkt[i], md - 2);
  }
  for (int i = 2; i <= SZ; i++) {
    if (!divide[i]) {
      for (int j = i; j <= SZ; j += i) {
        divide[j] = i;
      }
    }
  }
  while (t--) {
    cin >> k >> n;
    cout << getRes(n) << "\n";
  }
  return 0;
}