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 long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (!x) putchar('0');
  if (x < 0) x = -x, putchar('-');
  static int sta[20];
  register int tot = 0;
  while (x) sta[tot++] = x % 10, x /= 10;
  while (tot) putchar(sta[--tot] + 48);
}
const int MAXN = 1e6 + 5;
const int MOD = 1e9 + 7;
int n;
long long dp[2][22][2];
inline int get_num(int x, int y) { return n / (pow(2, x) * pow(3, y)); }
int main() {
  ios::sync_with_stdio(0);
  n = read();
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int t = 1, x = 0, y = 0;
  while ((t << 1) <= n) t <<= 1, x++;
  dp[1][x][0] = 1;
  if (t / 2 * 3 <= n) dp[1][x - 1][1] = 1, y = 1;
  for (int i = 1, cur = 1, nxt; i < n; ++i, cur ^= 1) {
    nxt = cur ^ 1;
    for (int j = 0; j <= x; ++j) {
      for (int k = 0; k <= y; ++k) {
        dp[nxt][j][k] = 0;
      }
    }
    for (int j = 0; j <= x; ++j) {
      for (int k = 0; k <= y; ++k) {
        if (dp[cur][j][k]) {
          int num = get_num(j, k);
          if (num < i) break;
          dp[nxt][j][k] =
              (dp[nxt][j][k] + dp[cur][j][k] * (num - i) % MOD) % MOD;
          if (j)
            dp[nxt][j - 1][k] =
                (dp[nxt][j - 1][k] +
                 dp[cur][j][k] * (get_num(j - 1, k) - num) % MOD) %
                MOD;
          if (k)
            dp[nxt][j][k - 1] =
                (dp[nxt][j][k - 1] +
                 dp[cur][j][k] * (get_num(j, k - 1) - num) % MOD) %
                MOD;
        }
      }
    }
  }
  cout << dp[n % 2][0][0] << endl;
  return 0;
}
