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 dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int Read() {
  char c;
  c = getchar();
  int tsqua = 0;
  while (c != ' ' && c != '\n') {
    tsqua = (((tsqua << 1) << 1) << 1) + (tsqua << 1) + (int)(c - '0');
    c = getchar();
  }
  return tsqua;
}
int n;
inline int GetCountOfDivisors(int x, int y) {
  return n / ((1 << x) * (y ? 3 : 1));
}
int DP[1000010][21][2];
int p;
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin >> n;
  while ((1 << p) <= n) p++;
  p--;
  DP[1][p][0] = 1;
  if ((1 << p - 1) * 3 <= n) DP[1][p - 1][1] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= p; j++)
      for (int k = 0; k < 2; k++) {
        if (i)
          DP[i + 1][j][k] =
              (DP[i + 1][j][k] +
               1ll * DP[i][j][k] * (GetCountOfDivisors(j, k) - i)) %
              1000000007;
        if (j)
          DP[i + 1][j - 1][k] =
              (DP[i + 1][j - 1][k] +
               1ll * DP[i][j][k] *
                   (GetCountOfDivisors(j - 1, k) - GetCountOfDivisors(j, k))) %
              1000000007;
        if (k)
          DP[i + 1][j][k - 1] =
              (DP[i + 1][j][k - 1] +
               1ll * DP[i][j][k] *
                   (GetCountOfDivisors(j, k - 1) - GetCountOfDivisors(j, k))) %
              1000000007;
      }
  cout << DP[n][0][0];
  return 0;
}
