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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const int N = 1000005, K = 20, mod = 1000000007;
int dp[N][K], f[N];
int main() {
  for (int i = 0; i < K; i++) dp[0][i] = 2;
  dp[0][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < K; j++) {
      for (int k = 0; k <= j; k++) {
        dp[i][j] =
            ((long long)dp[i][j] + dp[i - 1][k] + dp[i - 1][j - k]) % mod;
      }
      dp[i][j] = (long long)dp[i][j] * 500000004 % mod;
    }
  }
  for (int i = N - 1; i > 1; i--) {
    for (int j = i; j < N; j += i) {
      f[j] = i;
    }
  }
  int q = read();
  while (q--) {
    int r = read(), n = read(), ans = 1;
    while (n != 1) {
      int tt = f[n], sum = 0;
      while (n % tt == 0) {
        n /= tt;
        sum++;
      }
      ans = (long long)ans * dp[r][sum] % mod;
    }
    writeln(ans);
  }
}