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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
inline int readInt() {
  int x;
  scanf("%d", &x);
  return x;
}
long long dp[2][20][2];
long long f[20][2];
int main() {
  int n;
  cin >> n;
  int c2 = 0;
  int k = n;
  while (k >= 4) {
    c2++;
    k >>= 1;
  }
  int x = 1;
  for (int i = 0; i < (int)20; ++i) {
    f[i][0] = n / x;
    f[i][1] = n / (3 * x);
    x <<= 1;
  }
  if (k == 3) dp[1][c2][1] = 1;
  c2++;
  dp[1][c2][0] = 1;
  for (int i = 1; i < (int)n; ++i) {
    memset(dp[(i + 1) & 1], 0, sizeof(dp[(i + 1) & 1]));
    for (int j = 0; j < (int)c2 + 1; ++j) {
      for (int k = 0; k < (int)2; ++k) {
        if (dp[i & 1][j][k] == 0) continue;
        dp[(i + 1) & 1][j][k] =
            (dp[(i + 1) & 1][j][k] + (f[j][k] - i) * dp[i & 1][j][k]) %
            1000000007;
        if (j > 0)
          dp[(i + 1) & 1][j - 1][k] =
              (dp[(i + 1) & 1][j - 1][k] +
               max(f[j - 1][k] - f[j][k], 0LL) * dp[i & 1][j][k]) %
              1000000007;
        if (k > 0)
          dp[(i + 1) & 1][j][k - 1] =
              (dp[(i + 1) & 1][j][k - 1] +
               max(f[j][k - 1] - f[j][k], 0LL) * dp[i & 1][j][k]) %
              1000000007;
      }
    }
  }
  cout << dp[n & 1][0][0] << endl;
  return 0;
}
