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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int dp[N][22], sum[N][22], mi[N];
int main() {
  dp[0][0] = sum[0][0] = 1;
  for (int i = 1; i <= 21; ++i) sum[0][i] = sum[0][i - 1] + (dp[0][i] = 2);
  for (int i = 1; i <= N - 1; ++i) {
    dp[i][0] = sum[i][0] = 1;
    for (int j = 1; j <= 21; ++j)
      sum[i][j] = (sum[i][j - 1] + (dp[i][j] = sum[i - 1][j])) % P;
  }
  for (int i = 1; i <= N - 1; ++i) mi[i] = i;
  for (int i = 2; i <= N - 1; ++i)
    if (mi[i] == i) {
      for (int j = i; j < N; j += i) mi[j] = min(mi[j], i);
    }
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int ans = 1;
    while (n != 1) {
      int t = mi[n], k = 0;
      while (n % t == 0) n /= t, ++k;
      ans = (long long)ans * dp[r][k] % P;
    }
    printf("%d\n", ans);
  }
}