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 maxn = 2000007;
using ll = long long;
const ll mod = 1000000007;
const int D = 30;
int r, n;
ll way[D];
ll inv[maxn];
ll fac[maxn];
ll ifac[maxn];
int minp[maxn];
ll Get(ll a, ll b) {
  if (b == 0) return 0;
  return fac[a + b - 1] * ifac[a] % mod * ifac[b - 1] % mod;
}
int main() {
  fac[0] = ifac[0] = fac[1] = inv[1] = ifac[1] = 1;
  for (int i = 2; i < maxn; ++i) {
    inv[i] = (ll)mod / i * (mod - inv[mod % i]) % mod;
    fac[i] = (ll)fac[i - 1] * i % mod;
    ifac[i] = (ll)ifac[i - 1] * inv[i] % mod;
  }
  for (int i = 1; i < maxn; ++i) minp[i] = i;
  for (int i = 2; i < maxn; ++i)
    for (int j = i; j < maxn; j += i) minp[j] = min(minp[j], i);
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    if (n == 1) {
      printf("1\n");
      continue;
    }
    ll ans = 1;
    ll x = n;
    int t = 1, pre = minp[x];
    x /= minp[x];
    while (x > 1) {
      if (minp[x] != pre) {
        ans = (Get(t - 1, r + 1) * 2 % mod + Get(t, r)) % mod * ans % mod;
        t = 0;
      }
      pre = minp[x];
      x /= minp[x];
      t++;
    }
    if (t > 0)
      ans = (Get(t - 1, r + 1) * 2 % mod + Get(t, r)) % mod * ans % mod;
    printf("%I64d\n", ans);
  }
}