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 N = 1111111;
const int mod = 1000000007;
long long fac[N], ifac[N], inv[N];
int lpf[N];
inline long long modPow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) (ret *= a) %= mod;
    (a *= a) %= mod;
    b >>= 1;
  }
  return ret;
}
void pre() {
  fac[0] = ifac[0] = 1;
  fac[1] = inv[1] = ifac[1] = 1;
  for (int i = 2; i < N; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = (mod - ((mod / i) * inv[mod % i] % mod)) % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
  memset(lpf, -1, sizeof(lpf));
  for (int n = 2; n < N; n++) {
    if (lpf[n] == -1) {
      for (int m = n; m < N; m += n) {
        if (lpf[m] == -1) lpf[m] = n;
      }
    }
  }
}
inline long long ncr(int n, int r) {
  return (fac[n] * ifac[n - r] % mod) * ifac[r] % mod;
}
inline long long func(int a, int r) {
  if (r == 0) return 2;
  long long ret = ncr(a + r - 1, r) + ncr(a + r, r);
  if (ret >= mod) ret -= mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  pre();
  int q, n, r;
  cin >> q;
  while (q--) {
    cin >> r >> n;
    long long ans = 1;
    while (n > 1) {
      int cnt = 0;
      int x = lpf[n];
      while (n % x == 0) {
        n /= x;
        cnt++;
      }
      (ans *= func(cnt, r)) %= mod;
    }
    cout << ans << '\n';
  }
  return 0;
}