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>
int dp[(1000010)][(25)];
int leftPrime[(1000010)];
void preDp() {
  for (int i = 1; i < (25); ++i) dp[0][i] = 2;
  for (int i = 0; i < (1000010); ++i) dp[i][0] = 1;
  for (int i = 1; i < (1000010); ++i) {
    for (int j = 1; j < (25); ++j) {
      dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % (1000000007);
    }
  }
}
void preLeftPrime() {
  for (int i = 2; i < (1000010); ++i) {
    if (!leftPrime[i]) {
      for (int j = i; j < (1000010); j += i) {
        leftPrime[j] = i;
      }
    }
  }
}
int getans(int r, int n) {
  int res = 1;
  while (n != 1) {
    int cnt = 0, pri = leftPrime[n];
    while (!(n % pri)) {
      n /= pri;
      ++cnt;
    }
    res = (1ll) * res * dp[r][cnt] % (1000000007);
  }
  return res;
}
int main() {
  preDp();
  preLeftPrime();
  int q, r, n;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &r, &n);
    printf("%d\n", getans(r, n));
  }
  return 0;
}