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 int N = 1e6 + 7;
bool isPrime[N];
int prime[N / 10], cnt = 0, f[N];
inline void sieve() {
  memset(isPrime, 1, sizeof(isPrime));
  for (int i = 2; i < N; ++i) {
    if (isPrime[i]) prime[++cnt] = i, f[i] = i;
    for (int j = 1; j <= cnt && i * prime[j] < N; ++j) {
      isPrime[i * prime[j]] = 0;
      f[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
}
const long long MOD = 1e9 + 7;
int g[N][21], sum[N];
inline void init() {
  g[0][0] = 1;
  for (int i = 1; i <= 20; ++i) g[0][i] = 2;
  sum[0] = 1;
  for (int i = 1; i <= 20; ++i) sum[i] = sum[i - 1] + g[0][i];
  for (int t = 1; t < N; ++t) {
    g[t][0] = 1;
    for (int i = 1; i <= 20; ++i) g[t][i] = sum[i];
    sum[0] = 1;
    for (int i = 1; i <= 20; ++i) {
      sum[i] = sum[i - 1] + g[t][i];
      if (sum[i] >= MOD) sum[i] -= MOD;
    }
  }
}
int main() {
  sieve();
  init();
  int q, r, n;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    if (n == 1) {
      puts("1");
      continue;
    }
    long long ret = 1;
    while (n != 1) {
      int num = 0;
      int p = f[n];
      while (n % p == 0) num++, n /= p;
      ret *= 1LL * g[r][num];
      ret %= MOD;
    }
    printf("%I64d\n", ret);
  }
}