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 int P = 1000000007;
inline int ksm(int a, int t) {
  int r = 1;
  while (t) {
    if (t & 1) r = (long long)r * a % P;
    a = (long long)a * a % P;
    t >>= 1;
  }
  return r;
}
const int N = 1000005, M = 1000000;
int pri[N >> 3], pcnt, fl[N], fac[N * 2], ifac[N * 2], las[N];
void init() {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= 2 * M; i++) fac[i] = (long long)fac[i - 1] * i % P;
  ifac[2 * M] = ksm(fac[2 * M], P - 2);
  for (int i = 2 * M; i >= 2; i--) ifac[i - 1] = (long long)ifac[i] * i % P;
  for (int i = 2; i <= M; i++) {
    if (!fl[i]) pri[++pcnt] = i, las[i] = i;
    for (int j = 1; i * pri[j] <= M && j <= pcnt; j++) {
      fl[i * pri[j]] = 1;
      las[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
inline int f(int r, int k) {
  if (r == 0) return 2;
  int res = r + 2 * k;
  res = (long long)res * fac[r + k - 1] % P;
  res = (long long)res * ifac[r] % P;
  res = (long long)res * ifac[k] % P;
  return res;
}
int q, r, n, t;
int main() {
  init();
  int ans;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &r, &n);
    ans = 1;
    while (n > 1) {
      int p = las[n];
      t = 0;
      while (n % p == 0) n /= p, t++;
      ans = (long long)ans * f(r, t) % P;
    }
    printf("%d\n", ans);
  }
  return 0;
}