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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define mod 1000000007
#define inf 10000000000000007
#define eps 1e-13
#define PI 3.14159265359
#define sp printf(" ")
#define vi vector<int>
#define vl vector<long long>
#define vc vector<char>
#define vs vector<string>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pcc pair<char, char>
#define mp make_pair
#define FF first
#define SS second
#define pb(x) push_back(x)
#define sd(x) scanf("%d", &(x))
#define pd(x) printf("%d", (x))
#define pdn(x) printf("%d\n", (x))
#define sld(x) scanf("%ld", &(x))
#define pld(x) printf("%ld", (x))
#define pldn(x) printf("%ld\n", (x))
#define slf(x) scanf("%lf", &(x))
#define plf(x) printf("%.9lf", (x))
#define plfn(x) printf("%.9lf\n", (x))
#define sch(x) scanf("%c", &(x))
#define pch(x) printf("%c", (x))
#define pchn(x) printf("%c\n", (x))
#define gtl(x) getline(cin, (x))
#define flsh fflush(stdout)
#define ios ios::sync_with_stdio(false); cin.tie(0)
#define endl '\n'
using namespace std;
ll dp[1000004];
int main()
{
    int n;
    cin>>n;
    dp[0]=1;
    dp[1]=n;
    dp[2]=n*n;
    dp[2]%=mod;
    ll g=n-1;
    g=g*g;g=g%mod;
    ll sum=dp[1]+dp[2];
    sum%=mod;
    for(int i=3;i<=n;i++)
    {
        ll g=i-3;
        ll o=n-1;
        g=g*(o);
        g=g%mod;
        ll p=n-1-i+3;o++;
        p=p*o;p=p%mod;
        dp[i]=g+sum-dp[i-2]+p+mod;
        dp[i]%=mod;
        sum+=dp[i];
        sum%=mod;
     //   cout<<i<<" "<<dp[i]<<endl;

    }
    cout<<dp[n]<<endl;



}