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>
const int MAXN = 1e6, LOG = 20, Mod = 1e9 + 7;
int Add(int x, int y) {
  x += y;
  return x >= Mod ? x - Mod : x;
}
int Sub(int x, int y) {
  x -= y;
  return x < Mod ? x + Mod : x;
}
int Mul(int x, int y) { return 1ll * x * y % Mod; }
int prnum, prime[MAXN + 5], low[MAXN + 5];
bool vis[MAXN + 5];
void sieve() {
  low[1] = 1;
  vis[1] = 0;
  for (int i = 2; i <= MAXN; i++) {
    if (!vis[i]) prime[++prnum] = i, low[i] = i;
    for (int j = 1; j <= prnum && 1ll * i * prime[j] <= MAXN; j++) {
      vis[i * prime[j]] = 1;
      low[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
int f[MAXN + 5][LOG + 1];
void Init() {
  f[0][0] = 1;
  for (int i = 1; i <= LOG; i++) f[0][i] = 2;
  for (int i = 1; i <= MAXN; i++) {
    f[i][0] = f[i - 1][0];
    for (int j = 1; j <= LOG; j++)
      f[i][j] = Add(f[i][j], Add(f[i][j - 1], f[i - 1][j]));
  }
}
int T, r, n;
int main() {
  sieve();
  Init();
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &r, &n);
    int Ans = 1;
    while (n != 1) {
      int p = low[n], w = 0;
      for (; n % p == 0; n /= p, w++)
        ;
      Ans = Mul(Ans, f[r][w]);
    }
    printf("%d\n", Ans);
  }
}