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 Mod = 1e9 + 7;
long long f[1000006][25];
int pri[1000006], mn[1000006], tot = 0;
bool vis[1000006];
inline void init() {
  f[0][0] = 1;
  for (int k = 1; k <= 20; ++k) f[0][k] = 2;
  for (int r = 1; r <= 1e6; ++r) {
    f[r][0] = f[r - 1][0];
    for (int k = 1; k <= 20; ++k) {
      f[r][k] = (f[r][k - 1] + f[r - 1][k]) % Mod;
    }
  }
  for (int i = 2; i <= 1e6; ++i) {
    if (!vis[i]) pri[++tot] = i, mn[i] = i;
    for (int j = 1; j <= tot && i * pri[j] <= 1e6; ++j) {
      vis[pri[j] * i] = 1;
      mn[pri[j] * i] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
inline void work() {
  int r, n;
  scanf("%d%d", &r, &n);
  long long ans = 1;
  while (n != 1) {
    int t = mn[n], k = 0;
    while (mn[n] == t) n /= mn[n], ++k;
    (ans *= f[r][k]) %= Mod;
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  int m;
  scanf("%d", &m);
  while (m--) work();
  return 0;
}