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>
const int N = 1000005;
const int P = 1000000007;
using namespace std;
int f[2][N];
vector<int> v;
int calc(int n) {
  int i, j, k, cur = 0, prev;
  memset(f, 0, sizeof(f));
  f[0][0] = 1;
  for (i = 0; i < v.size(); i++) {
    prev = cur;
    cur = prev ^ 1;
    int p = n / v[i];
    int g = p;
    if (i > 0) g -= n / v[i - 1];
    f[cur][i + 1] = 1LL * f[prev][i] * g % P;
    for (j = i + 2; j <= p; j++) {
      f[cur][j] = 1LL * f[prev][j - 1] * g % P;
      f[cur][j] += 1LL * f[cur][j - 1] * (p - j + 1) % P;
      if (f[cur][j] >= P) f[cur][j] -= P;
    }
  }
  return f[cur][n];
}
int main() {
  int n, i, j;
  scanf("%d", &n);
  int p = 1, m = 0;
  while (p <= n) {
    p <<= 1;
    m++;
  }
  p >>= 1;
  m--;
  for (i = m; i >= 0; i--) {
    v.push_back(1 << i);
  }
  int ans = calc(n);
  if ((p + (p >> 1)) <= n) {
    for (i = 1; i <= m; i++) {
      v.clear();
      v.push_back(1);
      for (j = 1; j <= m; j++) {
        if (j == i)
          v.push_back(v[j - 1] * 3);
        else
          v.push_back(v[j - 1] << 1);
      }
      reverse(v.begin(), v.end());
      ans += calc(n);
      if (ans >= P) ans -= P;
    }
  }
  printf("%d\n", ans);
  return 0;
}
