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 N = 10000010;
const int mod = 1e9 + 7;
int n, M;
long long f[31][2], fac[N], inv[N], ans;
long long pow(long long x, int y) {
  long long ans = 1;
  for (; y; y >>= 1) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
  }
  return ans;
}
long long C(int x, int y) { return fac[x] * inv[y] % mod & inv[x - y] % mod; }
long long dfs(int left, int x, int y) {
  long long tmpf[31][2];
  for (int i = 0; i <= M; ++i)
    for (int j = 0; j <= 1; ++j) tmpf[i][j] = f[i][j];
  long long tot = 0;
  for (int i = x; i <= M; ++i)
    for (int j = y; j <= 1; ++j) tot += f[i][j], f[i][j] = 0;
  if (!x && !y) {
    for (int i = 0; i <= M; ++i)
      for (int j = 0; j <= 1; ++j) f[i][j] = tmpf[i][j];
    return fac[tot];
  }
  long long ans = 0;
  if (x) ans += dfs(left - tot, x - 1, y);
  if (y) ans += dfs(left - tot, x, y - 1);
  for (int i = 0; i <= M; ++i)
    for (int j = 0; j <= 1; ++j) f[i][j] = tmpf[i][j];
  return fac[left - 1] * inv[left - tot] % mod * tot % mod * ans % mod;
}
int main() {
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[n] = pow(fac[n], mod - 2);
  for (int i = n - 1; ~i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  f[0][0] = n;
  for (M = 1; f[M - 1][0]; ++M) f[M][0] = f[M - 1][0] >> 1;
  M -= 2;
  for (int i = 0; i <= M; ++i) f[i][0] -= f[i + 1][0];
  f[0][1] = n / 3;
  for (int i = 1; i <= M - 1; ++i) f[i][1] = f[i - 1][1] >> 1;
  for (int i = 0; i <= M - 1; ++i) f[i][1] -= f[i + 1][1];
  for (int i = 0; i <= M; ++i) f[i][0] -= f[i][1];
  ans = dfs(n, M, 0);
  if ((1 << (M - 1)) * 3 <= n) ans += dfs(n, M - 1, 1);
  printf("%lld\n", ans % mod);
  return 0;
}
