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;
using LL = long long;
const int MX = 1000006, mod = 1000000007;
int n, ans, d, pw2, num[23], dp[21][MX];
int solve1(int k, int d) {
  num[0] = n;
  for (int i = 1; i <= d; i++) num[i] = num[i - 1] >> 1;
  for (int i = 0; i < d; i++) num[i] -= num[i + 1];
  for (int i = d - 1; i >= 0; i--) num[i] += num[i + 1];
  for (int i = 0; i <= d; i++) memset(dp[i] + 1, 0, sizeof(int[num[i]]));
  dp[d][1] = 1;
  for (int i = d; i > 0; i--)
    for (int j = 1, val = num[i]; j <= val; j++)
      if (dp[i][j]) {
        dp[i][j + 1] = (dp[i][j + 1] + dp[i][j] * LL(val - j)) % mod;
        dp[i - 1][j + 1] =
            (dp[i - 1][j + 1] + dp[i][j] * LL(num[i - 1] - val)) % mod;
      }
  for (int j = 1; j <= n; j++)
    if (dp[0][j]) dp[0][j + 1] = (dp[0][j + 1] + dp[0][j] * LL(n - j)) % mod;
  return dp[0][n];
}
int f[23];
int solve2(int k, int d) {
  if (k > n || d == -1) return 0;
  long long ans = 0;
  for (int t = 0; t <= d; t++) {
    f[0] = k;
    for (int j = 1; j <= t; j++) f[j] = f[j - 1] >> 1;
    f[t + 1] = f[t] / 3;
    for (int j = t + 1; j <= d; j++) f[j + 1] = f[j] >> 1;
    for (int i = 0; i <= d + 1; i++) num[i] = n / f[i];
    for (int i = 0; i <= d + 1; i++) memset(dp[i] + 1, 0, sizeof(int[num[i]]));
    dp[0][1] = 1;
    for (int i = 0; i <= d; i++)
      for (int j = 1, val = num[i]; j <= val; j++)
        if (dp[i][j]) {
          dp[i][j + 1] = (dp[i][j + 1] + dp[i][j] * LL(val - j)) % mod;
          dp[i + 1][j + 1] =
              (dp[i + 1][j + 1] + dp[i][j] * LL(num[i + 1] - val)) % mod;
        }
    for (int j = 1; j <= n; j++)
      if (dp[d + 1][j])
        dp[d + 1][j + 1] = (dp[d + 1][j + 1] + dp[d + 1][j] * LL(n - j)) % mod;
    ans += dp[d + 1][n];
  }
  return ans % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (pw2 = 1; pw2 <= n; pw2 <<= 1, ++d)
    ;
  pw2 >>= 1, --d;
  cout << (solve1(pw2, d) + solve2(pw2 / 2 * 3, d - 1)) % mod;
  return 0;
}
