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 long long inf = 1LL << 61;
const long long mod = 1000000007;
const int maxn = 1000010;
const int maxf = 25;
int r, n;
long long dp[maxn][maxf];
int seiv[maxn];
vector<int> fa, fa_num;
int fa1(int m);
int main() {
  std::ios::sync_with_stdio(false);
  int i = 0, j = 0, k = 0;
  int sum;
  int q;
  scanf("%d", &q);
  for (i = 2; i < maxn; i++) {
    if (seiv[i]) continue;
    for (j = i; j < maxn; j += i) seiv[j] = i;
  }
  for (i = 0; i < maxn; i++) {
    for (j = 0; j < maxf; j++) {
      if (i == 0) {
        dp[i][j] = j == 0 ? 1 : 2;
      } else {
        if (j == 0) {
          dp[i][j] = 1;
        } else {
          dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
        }
      }
    }
  }
  while (q--) {
    scanf("%d%d", &r, &n);
    long long ans = 1;
    while (n > 1) {
      i = seiv[n];
      j = 0;
      while (n % i == 0) {
        n = n / i;
        j++;
      }
      ans = (ans * dp[r][j]) % mod;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
int fa1(int m) {
  int i = 2, sum = 0;
  fa.clear();
  fa_num.clear();
  for (; i * i <= m; i++) {
    if (m % i == 0) {
      sum = 0;
      while (m % i == 0) {
        sum++;
        m = m / i;
      }
      fa.push_back(i);
      fa_num.push_back(sum);
    }
  }
  if (m > 1) {
    fa.push_back(m);
    fa_num.push_back(1);
  }
  return sum;
}