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;
void In() { freopen("in.in", "r", stdin); }
void Out() { freopen("out.out", "w", stdout); }
const int N = 1e6 + 10;
const int M = 3e5 + 10;
const int Mbit = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int LP[N], f[N][25];
void init() {
  LP[1] = 1;
  for (int i = 2; i < N; i++)
    if (!LP[i])
      for (int j = i; j < N; j += i) LP[j] = i;
  for (int i = 0; i < N; i++) f[i][0] = 1;
  for (int i = 1; i < 25; i++) f[0][i] = 2;
  for (int i = 1; i < N; i++)
    for (int j = 1; j < 25; j++) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % mod;
}
long long solve(int r, int n) {
  long long ans = 1;
  while (n != 1) {
    int cnt = 0, p = LP[n];
    while (n % p == 0) n /= p, cnt++;
    ans = (ans * f[r][cnt]) % mod;
  }
  return ans;
}
int main() {
  init();
  int T, r, n;
  cin >> T;
  while (T--) {
    scanf("%d%d", &r, &n);
    printf("%lld\n", solve(r, n));
  }
  return 0;
}