Problem:
Score : 1000 points

Problem Statement

How many infinite sequences a_1, a_2, ... consisting of {{1, ... ,n}} satisfy the following conditions?

The n-th and subsequent elements are all equal. That is, if n \leq i,j, a_i = a_j.

For every integer i, the a_i elements immediately following the i-th element are all equal. That is, if i < j < k\leq i+a_i, a_j = a_k.

Find the count modulo 10^9+7.

Constraints

1 \leq n \leq 10^6

Input

Input is given from Standard Input in the following format:

n

Output

Print how many sequences satisfy the conditions, modulo 10^9+7.

Sample Input 1

2

Sample Output 1

4

The four sequences that satisfy the conditions are:

1, 1, 1, ...

1, 2, 2, ...

2, 1, 1, ...

2, 2, 2, ...

Sample Input 2

654321

Sample Output 2

968545283

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

#ifdef LOCAL111
	#define _GLIBCXX_DEBUG
#else
	#define NDEBUG
#endif
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
const int INF = 1e9;
using namespace std;
template<typename T, typename U> ostream& operator<< (ostream& os, const pair<T,U>& p) { cout << '(' << p.first << ' ' << p.second << ')'; return os; }

template<long long MOD>
class ModInt {
public:
	const static long long mod = MOD;
	long long x;
	
	ModInt() {
		x = 0;
	}

	ModInt(long long x) {
		this->x = x < 0 ? (x+mod)%mod : x%mod;
	}

	int get() const {
		return (int)x;
	}

	ModInt &operator+=(ModInt that) {
		if((x += that.get()) >= mod) x -= mod;
		return *this;
	}

	ModInt &operator-=(ModInt that) {
		if((x += mod-that.get()) >= mod) x -= mod;
		return *this;
	}

	ModInt &operator*=(ModInt that) {
		x = x*that.get()%mod;
		return *this;
	}

	ModInt &operator/=(ModInt that) {
		return *this *= that.inverse();
	}

	ModInt operator+(ModInt that) const {
		return ModInt(*this) += that;
	}

	ModInt operator-(ModInt that) const {
		return ModInt(*this) -= that;
	}

	ModInt operator*(ModInt that) const {
		return ModInt(*this) *= that;
	}

	ModInt operator/(ModInt that) const {
		return ModInt(*this) /= that;
	}

	ModInt inverse() const {
		using std::swap;
		long long a = x, b = mod, u = 1, v = 0;
		while(b) {
			long long t = a/b;
			a -= t*b; swap(a,b);
			u -= t*v; swap(u,v);
		}
		return ModInt(u);
	}

	ModInt pow(long long n){
		ModInt res = 1;
		ModInt x = *this;
		while(n != 0) {
			if(n&1){
				res *= x;
			}
			x *= x;
			n >>= 1;
		}
		return res;
	}

	bool operator==(ModInt that) const { return x == that.get(); }
	bool operator!=(ModInt that) const { return x != that.get(); };
	ModInt operator-() const { return x == 0 ? 0 : ModInt(mod-x); }

};

template<long long MOD> ostream& operator<< (ostream& os, const ModInt<MOD>& m) { cout << m.get(); return os; }
typedef ModInt<1000000007> mint;


#define endl '\n'
#define ALL(a)  (a).begin(),(a).end()
#define SZ(a) int((a).size())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n)  FOR(i,0,n)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#ifdef LOCAL111
	#define DEBUG(x) cout<<#x<<": "<<(x)<<endl
	template<typename T> void dpite(T a, T b){ for(T ite = a; ite != b; ite++) cout << (ite == a ? "" : " ") << *ite; cout << endl;}
#else
	#define DEBUG(x) true
	template<typename T> void dpite(T a, T b){ return; }
#endif
#define F first
#define S second
#define SNP string::npos
#define WRC(hoge) cout << "Case #" << (hoge)+1 << ": "
template<typename T> void pite(T a, T b){ for(T ite = a; ite != b; ite++) cout << (ite == a ? "" : " ") << *ite; cout << endl;}
template<typename T> bool chmax(T& a, T b){if(a < b){a = b; return true;} return false;}
template<typename T> bool chmin(T& a, T b){if(a > b){a = b; return true;} return false;}

typedef long long int LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;

void ios_init(){
	//cout.setf(ios::fixed);
	//cout.precision(12);
#ifdef LOCAL111
	return;
#endif
	ios::sync_with_stdio(false); cin.tie(0);	
}

mint bsolve(int n){
	mint res = 0;
	vector<int> a(n);
	vector<int> cnt(n);
	int hoge = 0;
	function<void(int)> rec = [&](int p){
		if(p == n){
			bool f = true;
			FOR(i,hoge,n){
				FOR(j,i+1,min(i+a[i]+1,n)){
					if(a[j] != a[i+1]){
						f = false;
					}
				}
			}
			if(f){
				dpite(ALL(a));
				res += 1;
			}
			return;
		}
		FOR(i,1,n+1){
			a[p] = i;
			rec(p+1);
		}
	};
	// REP(i,n){
	// 	res = 0;
	// 	hoge = i;
	// 	rec(i);
	// 	DEBUG(endl);
	// 	cnt[i] = res.get();
	// }
	hoge = 0;
	res = 0;
	DEBUG(23432);
	dpite(ALL(cnt));
	rec(0);
	return res;
}

int main()
{
	ios_init();
	int n;
	while(cin >> n) {
		vector<mint> dp(2*n+1,n);
		vector<mint> imos(2*n+1,0);
		dp.back() = n;
		RREP(i,2*n){
			imos[i] = imos[i+1]+dp[i];
		}
		RREP(i,n-1){
			dp[i] = dp[i+1];
			if(i+3 <= i+n+2){
				DEBUG(imos[i+3]-imos[i+n+2]);
				dp[i] += imos[i+3]-imos[i+n+2];
			}
			dp[i] += max(0ll,LL(n-i-3)*(n-1));
			// FOR(j,2,n+1){
			// 	int k = min(i+j+1,n-1);
			// 	dp[i] += dp[min(i+j+1,n-1)];
			// 	// dp[i] += n-1;
			// }
			imos[i] = imos[i+1]+dp[i];
		}
		dpite(ALL(dp));
		dpite(ALL(imos));
		// cout << bsolve(n) << endl;
		cout << dp[0] << endl;
	}
	return 0;
}