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 md = 1000000007;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
}
inline int mul(int a, int b) { return (long long)a * b % md; }
const int N = 1010101;
const int ALPHA = 21;
int f[N][ALPHA];
int p[N];
int a[N];
int dp[777][777];
int main() {
  for (int j = 0; j < ALPHA; j++) {
    f[0][j] = (j == 0);
  }
  for (int i = 1; i < N; i++) {
    int sum = 0;
    for (int j = 0; j < ALPHA; j++) {
      add(sum, f[i - 1][j]);
      f[i][j] = sum;
    }
  }
  for (int i = 1; i < N; i++) {
    p[i] = i;
  }
  for (int i = 2; i < N; i++) {
    if (p[i] == i) {
      for (int j = i + i; j < N; j += i) {
        if (p[j] == j) {
          p[j] = i;
        }
      }
    }
  }
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int r, n;
    scanf("%d %d", &r, &n);
    int cnt = 0;
    int tmp = n;
    while (tmp > 1) {
      int d = p[tmp];
      a[cnt] = 0;
      while (tmp % d == 0) {
        a[cnt]++;
        tmp /= d;
      }
      cnt++;
    }
    for (int i = 0; i <= cnt; i++) {
      for (int j = 0; j <= i; j++) {
        dp[i][j] = 0;
      }
    }
    dp[0][0] = 1;
    for (int i = 0; i < cnt; i++) {
      for (int j = 0; j <= i; j++) {
        add(dp[i + 1][j], mul(dp[i][j], f[r][a[i]]));
        add(dp[i + 1][j + 1], mul(dp[i][j], f[r + 1][a[i] - 1]));
      }
    }
    int ans = 0;
    for (int j = cnt; j >= 0; j--) {
      add(ans, ans);
      add(ans, dp[cnt][j]);
    }
    printf("%d\n", ans);
  }
  return 0;
}