Let's define a function f(p) on a permutation p as follows. Let g_i be the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of elements p_1, p_2, ..., p_i (in other words, it is the GCD of the prefix of length i). Then f(p) is the number of distinct elements among g_1, g_2, ..., g_n.

Let f_{max}(n) be the maximum value of f(p) among all permutations p of integers 1, 2, ..., n.

Given an integers n, count the number of permutations p of integers 1, 2, ..., n, such that f(p) is equal to f_{max}(n). Since the answer may be large, print the remainder of its division by 1000 000 007 = 10^9 + 7.

Input

The only line contains the integer n (2 ≤ n ≤ 10^6) — the length of the permutations.

Output

The only line should contain your answer modulo 10^9+7.

Examples

Input


2


Output


1

Input


3


Output


4

Input


6


Output


120

Note

Consider the second example: these are the permutations of length 3:

  * [1,2,3], f(p)=1. 
  * [1,3,2], f(p)=1. 
  * [2,1,3], f(p)=2. 
  * [2,3,1], f(p)=2. 
  * [3,1,2], f(p)=2. 
  * [3,2,1], f(p)=2. 



The maximum value f_{max}(3) = 2, and there are 4 permutations p such that f(p)=2.
#include <bits/stdc++.h>
using namespace std;
long long int LIM = 3e5 + 10;
long long int MOD = 1e9 + 7;
long long int n;
long long int exp(long long int n, long long int e) {
  if (e == 0) return 1;
  long long int temp = exp(n * n, e / 2);
  return e % 2 == 0 ? temp : temp * n;
}
long long int f(long long int x, long long int y) {
  return n / ((exp(2, x)) * exp(3, y));
}
int main() {
  cin >> n;
  long long int l = 0;
  long long int temp = 1;
  while (temp <= n) {
    temp = temp * 2;
    l++;
  }
  l--;
  long long int m = n / 3, g = 0;
  temp = 1;
  while (temp <= m) {
    temp = temp * 2;
    g++;
  }
  g--;
  long long int sum = max(g + 1, l);
  vector<vector<long long int> > store(l + 2, vector<long long int>(3, 0));
  for (int i = 0; i <= l; i++) {
    for (int j = 0; j <= 2; j++) store[i][j] = f(i, j);
  }
  vector<vector<vector<long long int> > > dp(
      2, vector<vector<long long int> >(l + 2, vector<long long int>(2, 0)));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 0; j <= l; j++) {
      for (long long int k = 0; k <= 1; k++) {
        if (i == 1) {
          if (j + k == sum && f(j, k) >= 1) dp[i % 2][j][k] = 1;
          continue;
        }
        long long int temp =
            ((long long int)dp[(i + 1) % 2][j][k] * (store[j][k] - i + 1)) %
            MOD;
        temp += ((long long int)dp[(i + 1) % 2][j + 1][k] *
                 ((store[j][k] - store[j + 1][k]) + MOD) % MOD) %
                MOD;
        if (k == 0)
          temp += ((long long int)dp[(i + 1) % 2][j][k + 1] *
                   ((store[j][k] - store[j][k + 1] + MOD) % MOD)) %
                  MOD;
        dp[i % 2][j][k] = temp % MOD;
      }
    }
  }
  cout << dp[n % 2][0][0] << endl;
  return 0;
}
