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 long long MOD = 1e9 + 7;
long long dp[2][21][2], p[21][2];
int main() {
  p[0][0] = 1, p[0][1] = 3;
  for (int i = 1; i < 20; ++i)
    p[i][0] = p[i - 1][0] << 1, p[i][1] = p[i][0] * 3;
  int n, m;
  scanf("%d", &n), m = log2(n);
  dp[1][m][0] = 1;
  if (p[m - 1][0] * 3 <= n) dp[1][m - 1][1] = 1;
  for (int i = 0; i <= m; ++i) p[i][0] = n / p[i][0], p[i][1] = n / p[i][1];
  for (int i = 2; i <= n; ++i)
    for (int j = m; j >= 0; --j)
      dp[i & 1][j][0] = ((p[j][0] - i + 1) * dp[!(i & 1)][j][0] +
                         (p[j][0] - p[j + 1][0]) * dp[!(i & 1)][j + 1][0] +
                         (p[j][0] - p[j][1]) * dp[!(i & 1)][j][1]) %
                        MOD,
             dp[i & 1][j][1] =
                 ((p[j][1] - i + 1) * dp[!(i & 1)][j][1] +
                  (p[j][1] - p[j + 1][1]) * dp[!(i & 1)][j + 1][1]) %
                 MOD;
  printf("%lld", dp[n & 1][0][0]);
  return 0;
}
