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;
const int N = 1e6 + 10;
const int B = 20;
const long long mod = 1e9 + 7;
int n;
long long pow2[B], pow3[2], f[2][B][2];
int main() {
  scanf("%d", &n);
  pow2[0] = 1;
  for (int i = 1; i < B; i++) pow2[i] = pow2[i - 1] * 2ll % mod;
  pow3[0] = 1, pow3[1] = 3;
  int num = 0;
  for (int bit = 0; bit < B; bit++)
    if ((1 << bit) & n) num = bit;
  f[1][num][0] = 1;
  if (num >= 1 && pow2[num - 1] * pow3[1] <= n) f[1][num - 1][1] = 1;
  for (int i = 2; i <= n; i++) {
    int now = i % 2, pre = now ^ 1;
    for (int j = 0; j < B; j++)
      for (int k = 0; k < 2; k++) {
        long long t = n / (pow2[j] * pow3[k]);
        long long a = t - i + 1;
        long long b = t - n / (pow2[j] * pow3[k] * 2);
        long long c = t - n / (pow2[j] * pow3[k] * 3);
        long long res = 0;
        if (a >= 0) res = (res + f[pre][j][k] * a % mod) % mod;
        if (b >= 0 && j + 1 < B) res = (res + f[pre][j + 1][k] * b % mod) % mod;
        if (c >= 0 && k + 1 < 2) res = (res + f[pre][j][k + 1] * c % mod) % mod;
        f[now][j][k] = res;
      }
  }
  printf("%lld\n", f[n % 2][0][0]);
  return 0;
}
