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;
constexpr int m = 1e9l + 7, N = 2e6;
int64_t mpow(int64_t a, int64_t b) {
  int64_t res = 1;
  for (int64_t p = a, d = 1; d <= b; d <<= 1, (p *= p) %= m)
    if (d & b) (res *= p) %= m;
  return res;
}
int64_t pow2(int64_t a) { return mpow(2, a); }
int64_t rev(int64_t a) { return mpow(a, m - 2); }
int64_t mul(int64_t a, int64_t b) { return a * b % m; }
int64_t divi(int64_t a, int64_t b) { return mul(a, rev(b)); }
int64_t sum(int64_t a, int64_t b) { return (a + b) % m; }
int64_t dif(int64_t a, int64_t b) { return (a - b + m) % m; }
int fac(int n) {
  static int f[N]{1};
  if (!f[n])
    for (int i = 1; i < N; ++i) f[i] = mul(f[i - 1], i);
  return f[n];
}
int rfac(int n) {
  static int rf[N]{1};
  if (!rf[n])
    for (int i = 1; i < N; ++i) rf[i] = mul(rf[i - 1], rev(i));
  return rf[n];
}
int C(int n, int k) { return mul(mul(fac(n), rfac(n - k)), rfac(k)); }
int S(int k, int n) { return C(n + k - 1, k); }
int Q(int r, int n) { return sum(S(r, n + 1), S(r, n)); }
int comp[N];
vector<int> primes;
int f(int r, int n) {
  int res = 1;
  for (auto i : primes)
    if (n % i == 0) {
      int a = 0;
      while (n % i == 0) n /= i, ++a;
      res = mul(res, Q(r, a));
    }
  if (n > 1) res = mul(res, Q(r, 1));
  return res;
}
int main() {
  for (int i = 2; i * i < N; ++i) {
    if (!comp[i]) {
      primes.push_back(i);
      for (int j = i * i; j < N; j += i) comp[j] = 1;
    }
  }
  int q, r, n;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    printf("%d\n", f(r, n));
  }
  return 0;
}