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>
using namespace std;
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define yes "YES"
#define no "NO"
#define F first
#define S second
#define sorta(a) sort(a.begin(),a.end())
#define all(a) (a.begin(),a.end())
#define forn(i,a,b) for(i=a;i<b;i++)
long long mod =1000000007;
vector<ll>prime(10000001,1);
vector<ll>sdf(10000001,1);
long long inf=1000000000000000007;
//ll C[101][101];


int isprime(ll n)
{
    ll i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
            {//cout<<i<<endl;
            return 0;}
    }
    return 1;
}
/*
int isperfectsq(ll n)
{


    long double x;
    x=(long double)(sqrt((long double )(n)));
    ll y=sqrt(n);
    if(x==y)
        return 1;
        else
            return 0;
}
*/
/*void ncr()
{
    C[0][0] = 1;
    for (int n = 1; n <= 500000; ++n) {
        C[n][0] = C[n][n] = 1;
        for (int k = 1; k < n; ++k)
            C[n][k] = C[n - 1][k - 1] + C[n - 1][k];
    }
}
*/
/*
void sieve()
{
    //cout<<"in1";
    ll i;
    //sdf[i]=1;
    ll j;
    prime[0]=prime[1]=0;
    for(i=2;i<=10000000;i++)
    {

            //isprime.push_back(i);
            for(j=i;j<=10000000;j+=i)
            {
                //prime[j]=0;

                sdf[j]++;
            }

    }

}
*/

long long modular_expo(long long x,long long y, long long mod)
{

    if(y==0)
        return 1;
    if(y%2==0)
        return modular_expo(((x%mod)*(x%mod))%mod,y/2,mod);
    if(y%2==1)
        return ((x%mod)*(modular_expo(((x%mod)*(x%mod))%mod,(y-1)/2,mod))%mod)%mod;


}
ll min3(ll a,ll b,ll c)
{
    ll m1=min(a,b);
    ll m2=min(a,c);
    return min(m1,m2);
}





int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);

    int i;
    ll n;
    cin>>n;

  vector<ll>fac(1000001);//

    fac[0]=fac[1]=1;
    //fac2[0]=fac2[1]=1;
    for(i=2;i<=1000000;i++)
    {
        fac[i]=(((fac[i-1])%mod)*(i%mod))%mod;
        //fac2[i]=fac2[i-1]*inv[i]%mod;
    }
    cout<<((fac[n]%mod)-(modular_expo(2,n-1,mod)%mod))%mod;


}


