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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
using namespace std;
inline int get() {
  int n = 0;
  char c;
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9') break;
    if (c == '-') goto s;
  }
  n = c - '0';
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9')
      n = n * 10 + c - '0';
    else
      return (n);
  }
s:
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9')
      n = n * 10 - c + '0';
    else
      return (n);
  }
}
int mindiv[1000001], prime[1000001];
unsigned char bv[1000001];
int f[1000001][20];
void print(int num) {
  if (num >= 10) print(num / 10);
  putchar(num % 10 + '0');
}
int main() {
  register int ptr = 0;
  for (register int i = 2; i <= 1000000; i++) {
    if (!bv[i]) {
      mindiv[i] = i;
      prime[ptr] = i;
      ptr++;
    }
    for (register int j = 0; j < ptr; j++) {
      int cjr = i * prime[j];
      if (cjr > 1000000) break;
      mindiv[cjr] = prime[j];
      bv[cjr] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  f[0][0] = 1;
  for (register int i = 1; i <= 19; i++) f[0][i] = 2;
  for (register int i = 1; i <= 1000000; i++) {
    long long sums = 0;
    for (register int j = 0; j <= 19; j++) {
      sums = (sums + f[i - 1][j]) % 1000000007;
      f[i][j] = sums;
    }
  }
  int t = get();
  while (t) {
    t--;
    int r = get(), n = get();
    long long ans = 1;
    while (n != 1) {
      int pri = mindiv[n];
      int t = 0;
      while (mindiv[n] == pri) {
        t++;
        n /= mindiv[n];
      }
      ans = (ans * f[r][t]) % 1000000007;
    }
    print(ans);
    putchar('\n');
  }
  return (0);
}