Problem:
Score : 800 points

Problem Statement

There are N squares lining up in a row, numbered 1 through N from left to right.
Initially, all squares are white.
We also have N-1 painting machines, numbered 1 through N-1.
When operated, Machine i paints Square i and i+1 black.

Snuke will operate these machines one by one.
The order in which he operates them is represented by a permutation of (1, 2, ..., N-1), P, which means that the i-th operated machine is Machine P_i.

Here, the score of a permutation P is defined as the number of machines that are operated before all the squares are painted black for the first time, when the machines are operated in the order specified by P.
Snuke has not decided what permutation P to use, but he is interested in the scores of possible permutations.
Find the sum of the scores over all possible permutations for him.
Since this can be extremely large, compute the sum modulo 10^9+7.

Constraints

2 \leq N \leq 10^6

Input

Input is given from Standard Input in the following format:

N

Output

Print the sum of the scores over all possible permutations, modulo 10^9+7.

Sample Input 1

4

Sample Output 1

16

There are six possible permutations as P.
Among them, P = (1, 3, 2) and P = (3, 1, 2) have a score of 2, and the others have a score of 3.
Thus, the answer is 2 \times 2 + 3 \times 4 = 16.

Sample Input 2

2

Sample Output 2

1

There is only one possible permutation: P = (1), which has a score of 1.

Sample Input 3

5

Sample Output 3

84

Sample Input 4

100000

Sample Output 4

341429644

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author majk
 */

#include <numeric>
#ifndef MAJK_LIB
#define MAJK_LIB

#include <vector>
#include <stack>
#include <iostream>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <fstream>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <string>
#include <sstream>
#include <queue>
#include <bitset>
using namespace std;

#define x first
#define y second
typedef std::pair<int,int> pii; typedef long long ll; typedef unsigned long long ull; typedef unsigned int ui; typedef pair<ui,ui> puu;

template <typename T, typename U> std::istream&operator>>(std::istream&i, pair<T,U>&p) {i >> p.x >> p.y; return i;}
template<typename T>std::istream&operator>>(std::istream&i,vector<T>&t) {for(auto&v:t){i>>v;}return i;}
template <typename T, typename U> std::ostream&operator<<(std::ostream&o, const pair<T,U>&p) {o << p.x << ' ' << p.y; return o;}
template<typename T>std::ostream&operator<<(std::ostream&o,const vector<T>&t) {if(t.empty())o<<'\n';for(size_t i=0;i<t.size();++i){o<<t[i]<<" \n"[i == t.size()-1];}return o;}
template<typename T> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T> bool in(T a, T b, T c) { return a <= b && b < c; }
ui logceil(int x) { return 8*sizeof(int)-__builtin_clz(x); }

namespace std { template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}}; }
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename F> double bshd(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){l=m;}else{h=m;}}return (l+h)/2;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename F> double bsld(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){h=m;}else{l=m;}}return (l+h)/2;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }

template<typename T>class vector2:public vector<vector<T>>{public:vector2(){} vector2(size_t a,size_t b,T t=T()):vector<vector<T>>(a,vector<T>(b,t)){}};
template<typename T>class vector3:public vector<vector2<T>>{public:vector3(){} vector3(size_t a,size_t b,size_t c,T t=T()):vector<vector2<T>>(a,vector2<T>(b,c,t)){}};
template<typename T>class vector4:public vector<vector3<T>>{public:vector4(){} vector4(size_t a,size_t b,size_t c,size_t d,T t=T()):vector<vector3<T>>(a,vector3<T>(b,c,d,t)){}};
template<typename T>class vector5:public vector<vector4<T>>{public:vector5(){} vector5(size_t a,size_t b,size_t c,size_t d,size_t e,T t=T()):vector<vector4<T>>(a,vector4<T>(b,c,d,e,t)){}};


#endif
#ifndef MOD_H
#define MOD_H



template <unsigned int N> class Field {
    typedef unsigned int ui;
    typedef unsigned long long ull;
	inline ui pow(ui a, ui p){ui r=1,e=a;while(p){if(p&1){r=((ull)r*e)%N;}e=((ull)e*e)%N;p>>=1;}return r;}
	/*extended GCD(slow):ll t=0,nt=1,r=N,nr=a;while(nr){ll q=r/nr;t-=q*nt;swap(t,nt);r-=q*nr;swap(r,nr);}assert(r<=1);return(t<0)?t+N:t;*/
	inline ui inv(ui a){return pow(a,N-2);}
public:
    inline Field(int x = 0) : v(x) {}
	inline Field<N> pow(int p){return (*this)^p; }
	inline Field<N> operator^(int p){return {(int)pow(v,(ui)p)};}
    inline Field<N>&operator+=(const Field<N>&o) {if ((ll)v+o.v >= N) v += o.v - N; else v += o.v; return *this; }
    inline Field<N>&operator-=(const Field<N>&o) {if (v<o.v) v -= o.v-N; else v-=o.v; return *this; }
    inline Field<N>&operator*=(const Field<N>&o) {v=(ull)v*o.v % N; return *this; }
    inline Field<N>&operator/=(const Field<N>&o) { return *this*=inv(o.v); }
    inline Field<N> operator+(const Field<N>&o) const {Field<N>r{*this};return r+=o;}
    inline Field<N> operator-(const Field<N>&o) const {Field<N>r{*this};return r-=o;}
    inline Field<N> operator*(const Field<N>&o) const {Field<N>r{*this};return r*=o;}
    inline Field<N> operator/(const Field<N>&o) const {Field<N>r{*this};return r/=o;}
    inline Field<N> operator-() {if(v) return {(int)(N-v)}; else return {0};};
    inline Field<N>& operator++() { ++v; if (v==N) v=0; return *this; }
    inline Field<N> operator++(int) { Field<N>r{*this}; ++*this; return r; }
    inline Field<N>& operator--() { --v; if (v==-1) v=N-1; return *this; }
    inline Field<N> operator--(int) { Field<N>r{*this}; --*this; return r; }
    inline bool operator==(const Field<N>&o) const { return o.v==v; }
	inline bool operator!=(const Field<N>&o) const { return o.v!=v; }
	inline explicit operator ui() const { return v; }
	inline static vector<Field<N>>fact(int t){vector<Field<N>>F(t+1,1);for(int i=2;i<=t;++i){F[i]=F[i-1]*i;}return F;}
	inline static vector<Field<N>>invfact(int t){vector<Field<N>>F(t+1,1);Field<N> X{1};for(int i=2;i<=t;++i){X=X*i;}F[t]=1/X;for(int i=t-1;i>=2;--i){F[i]=F[i+1]*(i+1);}return F;}
private: ui v;
};
template<unsigned int N>istream &operator>>(std::istream&is,Field<N>&f){unsigned int v;is>>v;f=v;return is;}
template<unsigned int N>ostream &operator<<(std::ostream&os,const Field<N>&f){return os<<(unsigned int)f;}
template<unsigned int N>Field<N> operator+(int i,const Field<N>&f){return Field<N>(i)+f;}
template<unsigned int N>Field<N> operator-(int i,const Field<N>&f){return Field<N>(i)-f;}
template<unsigned int N>Field<N> operator*(int i,const Field<N>&f){return Field<N>(i)*f;}
template<unsigned int N>Field<N> operator/(int i,const Field<N>&f){return Field<N>(i)/f;}


typedef Field<1000000007> FieldMod;

struct Ring {
	static int div(int p, int q, int N) {
		ll t=0,nt=1,r=N,nr=q;
		while(nr){ ll q=r/nr;t-=q*nt;r-=q*nr;swap(t,nt);swap(r,nr); }
		t=(t<0)?t+N:t;
		r=(r<0)?r+N:r;
		if (gcd(p,N)%r) { return 0; }
		return (int)(((ll)t*(ll)p/r)%N);
	}
};
#endif


class C {
public:
    void solve(istream& cin, ostream& cout) {
        int N; cin >> N;
        auto F = FieldMod::fact(N);
        auto I = FieldMod::invfact(N);

        FieldMod ans = F[N-1] * (N-1);

        FieldMod tot = 0;
        for (int i = 1; ; ++i) {
            if (N-2-i < i) break;

            ans -= F[N-2-i] * I[N-2-i-i] * F[N-1-i];

        }


        cout << ans << endl;

//        int ans = 0;
//        vector<int> P(N-1);
//        iota(P.begin(),P.end(),0);
//        do {
//            vector<bool> S(N, true);
//            int c = N;
//            for (int i = 0; i < N-1; ++i) {
//                c -= S[P[i]];
//                c -= S[P[i]+1];
//                S[P[i]] = false;
//                S[P[i]+1] = false;
//                if (c == 0) { ans += i+1; break; }
//            }
//        } while (next_permutation(P.begin(),P.end()));
//        cout << ans << '\n';
    }
};


int main() {
	ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	C solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
    return 0;
}