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 = 1000000 + 5;
const int mod = 1e9 + 7;
int q, r, n;
int pri[N], cnt;
int mindiv[N];
long long f[N][22], sum[22];
bool vis[N];
void rd(int &x) {
  x = 0;
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  for (x = (ch ^ '0'); isdigit(ch = getchar());
       x = (x << 1) + (x << 3) + (ch ^ '0'))
    ;
}
void sieve() {
  mindiv[1] = 1;
  for (int i = 2; i <= N - 3; i++) {
    if (!vis[i]) {
      pri[++cnt] = i;
      mindiv[i] = i;
    }
    for (int j = 1; j <= cnt; j++) {
      if (pri[j] * i > N - 3) break;
      vis[pri[j] * i] = 1;
      mindiv[pri[j] * i] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  sieve();
  f[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= 19; i++) f[0][i] = 2, sum[i] = sum[i - 1] + f[0][i];
  for (register int i = 1; i <= N - 3; i++) {
    for (int j = 0; j <= 19; j++) {
      f[i][j] = sum[j];
      sum[j] = 0;
      if (j) sum[j] = sum[j - 1];
      (sum[j] += f[i][j]) %= mod;
    }
  }
  int t;
  rd(t);
  while (t--) {
    rd(r), rd(n);
    long long ans = 1;
    while (n != 1) {
      long long div = mindiv[n];
      int cnt = 0;
      while (mindiv[n] == div) cnt++, n /= mindiv[n];
      (ans *= f[r][cnt]) %= mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}