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>
const int P = 1000000007;
int q, r, n, np[2004], cnt, p[999];
int fac[1000100], inv[1000100], ifac[1000100];
int C(int n, int m) {
  if (n < m || m < 0) return 0;
  return (long long)fac[n] * ifac[n - m] % P * ifac[m] % P;
}
int main() {
  n = 2e3;
  for (int i = 2; i <= n; i++) {
    if (!np[i]) p[++cnt] = i;
    for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
      np[i * p[j]] = true;
      if (i % p[j] == 0) break;
    }
  }
  n = 1e6 + 50;
  fac[0] = ifac[0] = inv[1] = ifac[1] = fac[1] = 1;
  for (int i = 2; i <= n; i++) {
    fac[i] = (long long)fac[i - 1] * i % P;
    inv[i] = (long long)inv[P % i] * (P - P / i) % P;
    ifac[i] = (long long)ifac[i - 1] * inv[i] % P;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    int s = 1;
    for (int j = 1; j <= cnt && p[j] * p[j] <= n; j++)
      if (n % p[j] == 0) {
        int k = 0;
        while (n % p[j] == 0) n /= p[j], ++k;
        int f = (2 * C(r + 1 + k - 1, r) - C(r + k - 1, r - 1)) % P;
        s = (long long)s * f % P;
      }
    if (n > 1) s = (long long)s * (r + 2) % P;
    s = (s + P) % P;
    printf("%d\n", s);
  }
  return 0;
}