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 = 1000005;
const int mod = 1000000007;
void add(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int d[21][N];
int lp[N], pr[N], len = 0;
void precalc() {
  for (int r = 0; r <= 1000000; ++r) d[0][r] = 1;
  for (int i = 1; i < 20; ++i) d[i][0] = 2;
  for (int r = 1; r <= 1000000; ++r)
    for (int i = 1; i < 20; ++i) {
      d[i][r] = d[i - 1][r];
      add(d[i][r], d[i][r - 1]);
    }
  for (int i = 2; i <= 1000000; ++i) {
    if (!lp[i]) {
      lp[i] = i;
      pr[++len] = i;
    }
    for (int j = 1; j <= len && i * 1ll * pr[j] <= 1000000 && pr[j] <= lp[i];
         ++j)
      lp[pr[j] * i] = pr[j];
  }
}
inline int calc(int &r, int &n) {
  int res = 1;
  while (n > 1) {
    int x = lp[n], cnt = 0;
    while (n % x == 0) {
      n /= x;
      ++cnt;
    }
    res = res * 1ll * d[cnt][r] % mod;
  }
  return res;
}
int main() {
  precalc();
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    printf("%d\n", calc(r, n));
  }
  return 0;
}