A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Consider a permutation p of length n, we build a graph of size n using it as follows: 

  * For every 1 ≤ i ≤ n, find the largest j such that 1 ≤ j < i and p_j > p_i, and add an undirected edge between node i and node j 
  * For every 1 ≤ i ≤ n, find the smallest j such that i < j ≤ n and p_j > p_i, and add an undirected edge between node i and node j 



In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.

For clarity, consider as an example n = 4, and p = [3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).

A permutation p is cyclic if the graph built using p has at least one simple cycle. 

Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 10^9+7.

Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3 ≤ n ≤ 10^6).

Output

Output a single integer 0 ≤ x < 10^9+7, the number of cyclic permutations of length n modulo 10^9+7.

Examples

Input


4


Output


16

Input


583291


Output


135712853

Note

There are 16 cyclic permutations for n = 4. [4,2,1,3] is one such permutation, having a cycle of length four: 4 → 3 → 2 → 1 → 4.

Nodes v_1, v_2, …, v_k form a simple cycle if the following conditions hold: 

  * k ≥ 3. 
  * v_i ≠ v_j for any pair of indices i and j. (1 ≤ i < j ≤ k) 
  * v_i and v_{i+1} share an edge for all i (1 ≤ i < k), and v_1 and v_k share an edge. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define time__(d) for ( auto blockTime = make_pair(chrono::high_resolution_clock::now(), true);blockTime.second;\
debug("%s: %lld ms\n", d, chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - blockTime.first).count()), blockTime.second = false)
#define li long long int
#define ld long double
#define eb push_back
#define pb pop_back
#define INF 1e18+10
#define MINF -1e18-10
#define rep(i,a,b) for(li i=a;i<b;i++)
#define size(a) a.size()
#define rev(a) reverse(a.begin(),a.end())
#define len(a) a.length()
#define f first
#define s second
#define sort(a) sort(a.begin(),a.end())
#define vli vector<li>
#define vvli vector<vli >
#define pli pair<li,li>
#define vpli vector<pli>
#define inp(t) li t;cin>>t; while(t--)
using namespace std;
const li mod = 1e9+7;
#define MAXN 1000005 

li power(li a, li b) { a%=mod; li ret=1;while(b) {if(b&1) ret*=a;a*=a;if(ret>=mod) ret%=mod;if(a>=mod) a%=mod;b>>=1;}return ret;}

bool isprime(li n)
{
  if (n == 1)
    return false;
  int i = 2;
  while (i*i <= n){
      if (n % i == 0)return false;
      i += 1;
  }
  return true;
}

li lcm(li a,li b)
{
    return ((a*b)/__gcd(a,b));
}

void solve()
{
    li n;
    cin>>n;

    li fact = 1;
    for(li i=1;i<=n;i++)
      fact = (fact*i)%mod;

    li powerterm = power(2,n-1);

    cout<<(fact%mod - powerterm%mod)%mod<<endl;
}

int main()
{
    IOS;
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif

    solve();
    return 0;
}