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;
int prime[1000050][20];
int cnt[1000050];
int dp[1000005][30];
void init() {
  int i, j;
  for (i = 2; i <= 1000000; i++) {
    if (cnt[i]) continue;
    prime[i][0] = i;
    cnt[i] = 1;
    for (j = 2; i * j <= 1000000; j++) prime[i * j][cnt[i * j]++] = i;
  }
  for (int i = 1; i < 30; i++) dp[0][i] = 2;
  for (int i = 1; i <= 1000000; i++) dp[i][0] = 1;
  dp[0][0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    for (int j = 1; j < 30; j++) {
      dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % (1000000007);
    }
  }
}
int main() {
  int q;
  cin >> q;
  init();
  while (q--) {
    int r, n;
    scanf("%d %d", &r, &n);
    long long ans = 1;
    for (int i = 0; i < cnt[n]; i++) {
      int temp = prime[n][i];
      int counter = 0;
      int tem = n;
      while (tem % temp == 0) {
        counter++;
        tem /= temp;
      }
      ans = (ans * (long long)dp[r][counter]) % (long long)(1000000007);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}