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;
double PI = acos(-1);
double EPS = 1e-6;
int cmp(double a, double b) {
  return (a + EPS < b) ? -1 : (a - EPS > b) ? 1 : 0;
}
bool v[1000005];
int len, sp[1000005];
long long MOD = 1e9 + 7;
void Sieve() {
  for (long long i = 2; i < 1000005; i += 2) sp[i] = 2;
  for (long long i = 3; i < 1000005; i += 2) {
    if (!v[i]) {
      sp[i] = i;
      for (long long j = i; (j * i) < 1000005; j += 2) {
        if (!v[j * i]) v[j * i] = true, sp[j * i] = i;
      }
    }
  }
}
long long dp[1000023][22];
void pre(int n, int m) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (j == 0)
        dp[i][0] = 1;
      else if (j > i)
        dp[i][j] = 0;
      else
        dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % MOD;
    }
  }
}
long long fsolve(long long n, long long p) {
  if (n == 0) return 2;
  if (p == 0) return 1;
  long long ans = (dp[n + p - 1][p - 1] * 2) % MOD;
  ans = (ans + dp[n + p - 1][p]) % MOD;
  return ans;
}
long long solve(long long lvl, long long N) {
  long long ans = 1;
  while (N != 1) {
    long long np = sp[N], ex = 0;
    while (N % np == 0) {
      ex++;
      N /= np;
    }
    ans = (ans * fsolve(lvl, ex)) % MOD;
  }
  return ans;
}
int main() {
  int k, N;
  pre(1000023, 22);
  Sieve();
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &k, &N);
    printf("%I64d\n", solve(k, N));
  }
}