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;
long long g[31][31], f[31], a[31], b[31], fac[10000010], inv[10000010];
inline long long C(int m, int n) {
  if (n < m) return 0;
  if (n == m) return 1;
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
inline long long dfs(int x, int y, int x1, int y1, int n) {
  if (x < 0 || y < 0) return 0;
  if (!x && !y) return fac[n];
  long long w = g[x][y] - g[x1][y1];
  return fac[w] * C(w - 1, n - 1) % 1000000007 *
         (dfs(x - 1, y, x, y, n - w) + dfs(x, y - 1, x, y, n - w)) % 1000000007;
}
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int main() {
  int n = read();
  inv[0] = inv[1] = fac[0] = fac[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = inv[1000000007 % i] * (1000000007 - 1000000007 / i) % 1000000007;
  for (int i = 2; i <= n; i++)
    fac[i] = fac[i - 1] * i % 1000000007,
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  int m = log2(n);
  for (int i = 0; i <= m; i++)
    g[i][0] = n / (1 << i), g[i][1] = n / ((1 << i) * 3);
  long long ans = dfs(m, 0, 30, 30, n);
  if ((1 << (m - 1)) * 3 <= n) ans += dfs(m - 1, 1, 30, 30, n);
  cout << ans % 1000000007 << endl;
  return 0;
}
