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: Time Limit Exceeded)

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define FOR(i,n,m) for(int i=(int)(n); i<=(int)(m); i++)
#define RFOR(i,n,m) for(int i=(int)(n); i>=(int)(m); i--)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

//-------------------------------------------------
//--ModInt
//-------------------------------------------------
const ::std::uint_fast64_t MOD = 1e9+7;

class mint
{
private:
    using value_type = ::std::uint_fast64_t;
    value_type n;
public:
    mint():n(0){}
    mint(::std::int_fast64_t _n):n(_n<0 ? MOD-(-_n)%MOD : _n%MOD){}
    mint(const mint &m):n(m.n){}

    friend ::std::ostream& operator<<(::std::ostream &os, const mint &a){
		return os << a.n;
	}
    friend ::std::istream& operator>>(::std::istream &is, mint &a){
        value_type temp; is>>temp;
        a = mint(temp);
        return is;
    }

    mint& operator+=(const mint &m){n+=m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator-=(const mint &m){n+=MOD-m.n; n=(n<MOD)?n:n-MOD; return *this;}
    mint& operator*=(const mint &m){n=n*m.n%MOD; return *this;}
    mint& operator/=(const mint &m){return *this*=m.inv();}
    mint& operator++(){return *this+=1;}
    mint& operator--(){return *this-=1;}

    mint operator+(const mint &m) const {return mint(*this)+=m;}
    mint operator-(const mint &m) const {return mint(*this)-=m;}
    mint operator*(const mint &m) const {return mint(*this)*=m;}
    mint operator/(const mint &m) const {return mint(*this)/=m;}
    mint operator++(int){mint t(*this); *this+=1; return t;}
    mint operator--(int){mint t(*this); *this-=1; return t;}

    bool operator==(const mint &m) const {return n==m.n;}
    bool operator!=(const mint &m) const {return n!=m.n;}
    mint operator-() const {return mint(MOD-n);}

    mint pow(value_type b) const {
        mint ret(1), m(*this);
        while(b){
            if (b & 1) ret*=m;
            m*=m;
            b>>=1;
        }
        return ret;
    }

    mint inv() const {return pow(MOD-2);}
};

//-------------------------------------------------
//--Lazy Evaluation Segment tree (Proportional)
//-------------------------------------------------
template<typename T, typename E>
class LazySegmentTree
{
private:
    using F = ::std::function<T(T,T)>;
    using G = ::std::function<T(T,E)>;
    using H = ::std::function<E(E,E)>;
    F f; G g; H h;
    int n, height;
    ::std::vector<T> seg;
    ::std::vector<E> lazy;
    T te; E ee;
public:
    LazySegmentTree(int _n, F f, G g, H h, T te, E ee)
        : f(f),g(g),h(h),te(te),ee(ee)
    {
        init(_n);
    }
    LazySegmentTree(::std::vector<T> &v, F f, G g, H h, T te, E ee)
        : f(f),g(g),h(h),te(te),ee(ee)
    {
        int _n = v.size();
        init(_n);
        for(int i=n; i<n+_n; i++) seg[i] = v[i-n];
        build();
    }
    void init(int _n){
        n = 1; height=0;
        while(n<_n) n<<=1, height++;
        seg.resize(2*n, te);
        lazy.resize(2*n, ee);
    }
    void build(){
        for(int i=n-1; i>0; i--) seg[i] = f(seg[i<<1], seg[i<<1|1]);
    }
    T eval(int k){
        int c = breadth(k);
        return (lazy[k]==ee) ? seg[k] : g(seg[k],quantify(lazy[k], c));
    }
    void waveto(int k){
        for(int i=height; i>0; i--) wave(k>>i);
    }
    void wave(int k){
        seg[k] = eval(k);
        lazy[k<<1] = h(lazy[k<<1],lazy[k]);
        lazy[k<<1|1] = h(lazy[k<<1|1],lazy[k]);
        lazy[k] = ee;
    }
    void recalc(int k){
        for(int i=1; i<=height; i++){
            int m = k>>i;
            seg[m] = f(eval(m<<1), eval(m<<1|1));
        }
    }
    void update(int a, int b, E x){
        if (b<=a) return;
        waveto(a+=n);
        waveto(b+=n-1);
        T L=ee, R=ee;
        int l=a, r=b;
        while(l<=r){
            if (l&1) lazy[l] = h(lazy[l], x), l++;
            if (!(r&1)) lazy[r] = h(lazy[r], x), r--;
            l>>=1; r>>=1;
        }
        recalc(a);
        recalc(b);
    }
    T query(int a, int b){
        waveto(a+=n);
        waveto(b+=n-1);
        T L=te, R=te;
        int l=a, r=b;
        while(l<=r){
            if (l&1) L = f(L, eval(l++));
            if (!(r&1)) R = f(eval(r--), R);
            l>>=1; r>>=1;
        }
        return f(L,R);
    }
    T operator[](int k){return query(k, k+1);};
    inline int breadth(int k){
        int ret = height-31+__builtin_clz(k);
        return 1<<ret;
    }
    //return value proportional to segment's width
    inline E quantify(E laz, int c){
        return laz*c;
    }
};

//-------------------------------------------------

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin>>n;
    if (n==1){
        cout<<"1\n";
        return 0;
    }
    int N = 1000000;
    LazySegmentTree<mint,mint> seg(N,
        [](mint x, mint y){return x+y;},
        [](mint x, mint y){return x+y;},
        [](mint x, mint y){return x+y;},
        0,0
    );
    seg.update(0,1,1);
    seg.update(3,N,1);
    FOR(i,1,N){
        mint s = seg[i-1];
        seg.update(i,i+1,s);
        mint t = seg[i];
        seg.update(i+3,N,t);
    }
    mint ans = seg.query(0,n-1)*(n*(n-1)-n+1);
    ans+=seg.query(0,n)*(n-1);
    ans+=1;
    cout<<ans<<"\n";
    return 0;
}