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 N = 1010101;
const int MOD = 1000000007;
const int INV2 = (MOD + 1) / 2;
inline char get(void) {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template <typename T>
inline void read(T &x) {
  static char c;
  x = 0;
  for (c = get(); c < '0' || c > '9'; c = get())
    ;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}
int n, r, q, x, sum, Pcnt, ans, lim;
vector<int> fact[N];
int f[N][30];
int prime[N], vis[N];
inline void Add(int &x, int a) {
  x += a;
  while (x >= MOD) x -= MOD;
}
int main(void) {
  read(q);
  f[0][0] = 1;
  lim = 1000000;
  for (int i = 1; i <= 20; i++) f[0][i] = 2;
  for (int i = 1; i <= lim; i++) {
    sum = 0;
    for (int j = 0; j <= 20; j++) {
      Add(sum, f[i - 1][j]);
      Add(f[i][j], sum);
    }
  }
  for (int i = 2; i <= lim; i++) {
    if (!vis[i]) prime[++Pcnt] = i;
    for (int j = 1; j <= Pcnt && (x = prime[j] * i) <= lim; j++) {
      vis[x] = 1;
      if (!(i % prime[j])) break;
    }
  }
  for (int i = 1; i <= Pcnt; i++)
    for (int j = prime[i]; j <= lim; j += prime[i]) fact[j].push_back(prime[i]);
  while (q--) {
    read(n);
    read(r);
    ans = 1;
    for (int d : fact[r]) {
      x = 0;
      while (!(r % d)) {
        r /= d;
        ++x;
      }
      ans = (long long)ans * f[n][x] % MOD;
    }
    printf("%d\n", ans);
  }
  return 0;
}