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 + 1, P = 1e9 + 7;
int i, j, k, n, q, r, d[MAXN], f[MAXN][20];
int main() {
  for (i = 2; i < MAXN; i++)
    if (!d[i])
      for (j = i; j < MAXN; j += i) d[j] = i;
  for (i = 0; i < MAXN; *f[i++] = 1)
    ;
  for (d[1] = i = 1; i < 20; f[0][i++] = 2)
    ;
  for (i = 1; i < MAXN; i++)
    for (j = 1; j < 20; j++) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % P;
  for (scanf("%d", &q); q--; printf("%d\n", r)) {
    scanf("%d%d", &k, &n);
    for (r = 1; n > 1; r = 1ll * r * f[k][i] % P)
      for (j = d[n], i = 0; n % j == 0; n /= j) i++;
  }
}