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 MOD = 1000 * 1000 * 1000 + 7;
const int MAXN = 1000000;
int dp[MAXN + 1][21], sum[MAXN + 1][21], p[MAXN + 1];
void add(int& x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  dp[0][0] = 1;
  sum[0][0] = 1;
  for (int i = 1; i <= 20; i++) {
    dp[0][i] = 2;
    sum[0][i] = 2;
    if (i > 0) add(sum[0][i], sum[0][i - 1]);
  }
  for (int r = 1; r <= MAXN; r++) {
    for (int n = 0; n <= 20; n++) {
      dp[r][n] = sum[r - 1][n];
      sum[r][n] = dp[r][n];
      if (n > 0) add(sum[r][n], sum[r][n - 1]);
    }
  }
  for (int i = 2; i <= MAXN; i++) {
    if (!p[i]) {
      p[i] = i;
      for (int j = i + i; j <= MAXN; j += i) p[j] = i;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int r, n;
    scanf("%d %d", &r, &n);
    int ans = 1;
    while (n > 1) {
      int d = p[n];
      int cnt = 0;
      while (n % d == 0) cnt++, n /= d;
      ans = ((long long)ans * dp[r][cnt]) % MOD;
    }
    printf("%d\n", ans);
  }
  return 0;
}