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 q, r, n;
int s[1000007];
vector<int> p[1000007];
long long dp[1000007][22];
void dpCal() {
  for (int i = 0; i <= 20; ++i) dp[0][i] = 2;
  for (int i = 0; i <= 1000000; ++i) dp[i][0] = 1;
  for (int i = 1; i <= 1000000; ++i) {
    for (int k = 1; k <= 20; ++k) {
      dp[i][k] = (dp[i - 1][k] + dp[i][k - 1]) % 1000000007LL;
    }
  }
}
int getLog(int n, int p) {
  int ans = 0;
  while (n % p == 0) {
    n /= p;
    ans += 1;
  }
  return ans;
}
void factor() {
  for (int i = 1; i <= 1000000; ++i) {
    p[i].clear();
    s[i] = 1;
  }
  for (int i = 2; i <= 1000000; ++i) {
    if (s[i] == 1) {
      for (int j = i; j <= 1000000; j += i) {
        p[j].push_back(getLog(j, i));
        s[j] = 0;
      }
    }
  }
}
int main() {
  dpCal();
  factor();
  scanf("%d", &q);
  long long ans;
  while (q--) {
    scanf("%d%d", &r, &n);
    ans = 1;
    for (auto k : p[n]) {
      ans = (ans * dp[r][k]) % 1000000007LL;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}