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;
namespace STDquantum {
template <typename T>
inline void read(T &x) {
  char ch;
  while (ch = getchar(), ch > '9' || ch < '0') {
  }
  x = (ch ^ 48);
  while (ch = getchar(), ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48);
}
template <typename T>
inline void write(T x) {
  static int stk[100], top = 0;
  if (x == 0) return (void)putchar('0');
  if (x < 0) x = -x, putchar('-');
  while (x) stk[++top] = x % 10, x /= 10;
  while (top) putchar(stk[top--] + '0');
}
const int N = 1e6 + 10, K = 20, mod = 1e9 + 7;
int low[N];
void getPrime() {
  for (int i = 2; i < N; ++i) {
    if (!low[i]) {
      for (int j = i; j < N; j += i) {
        low[j] = i;
      }
    }
  }
}
int dp[N][K], sum[K] = {1};
void init() {
  getPrime();
  for (int i = 0; i < N; ++i) {
    dp[i][0] = 1;
  }
  for (int i = 1; i < K; ++i) dp[0][i] = 2, sum[i] = sum[i - 1] + dp[0][i];
  for (int i = 1; i < N; ++i) {
    for (int j = 1; j < K; ++j) {
      dp[i][j] = sum[j];
      sum[j] = (sum[j - 1] + dp[i][j]) % mod;
    }
  }
}
int r, n, q;
long long ans = 1;
inline void main() {
  init();
  for (read(q); q--; ans = 1) {
    read(r), read(n);
    int cnt, p;
    while (n != 1) {
      cnt = 0, p = low[n];
      while (n % p == 0) ++cnt, n /= p;
      ans = ans * dp[r][cnt] % mod;
    }
    write(ans), putchar('\n');
  }
}
}  // namespace STDquantum
int main() {
  STDquantum::main();
  return 0;
}