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.

Examples

Input

2


Output

4


Input

654321


Output

968545283
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;

public class Main {
    static class Scanner {
        BufferedReader br;
        StringTokenizer tk = new StringTokenizer("");

        public Scanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public int nextInt() throws IOException {
            if (tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
            tk = new StringTokenizer(br.readLine());
            return nextInt();
        }

        public long nextLong() throws IOException {
            if (tk.hasMoreTokens())
                return Long.parseLong(tk.nextToken());
            tk = new StringTokenizer(br.readLine());
            return nextLong();
        }

        public String next() throws IOException {
            if (tk.hasMoreTokens())
                return (tk.nextToken());
            tk = new StringTokenizer(br.readLine());
            return next();
        }

        public String nextLine() throws IOException {
            tk = new StringTokenizer("");
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            if (tk.hasMoreTokens())
                return Double.parseDouble(tk.nextToken());
            tk = new StringTokenizer(br.readLine());
            return nextDouble();
        }

        public char nextChar() throws IOException {
            if (tk.hasMoreTokens())
                return (tk.nextToken().charAt(0));
            tk = new StringTokenizer(br.readLine());
            return nextChar();
        }

        public int[] nextIntArray(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public long[] nextLongArray(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArrayOneBased(int n) throws IOException {
            int a[] = new int[n + 1];
            for (int i = 1; i <= n; i++)
                a[i] = nextInt();
            return a;
        }

        public long[] nextLongArrayOneBased(int n) throws IOException {
            long a[] = new long[n + 1];
            for (int i = 1; i <= n; i++)
                a[i] = nextLong();
            return a;
        }


    }

    public static void main(String args[]) throws IOException {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }, "1", 1 << 26).start();

    }

    static void solve() throws IOException {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n=in.nextInt();
        long dp[]=new long[2*n+5];
        dp[n]=n;
        for(int i=n+1;i<=2*n+4;i++)
            dp[i]=1;
        int st=n+3;
        int en=n+n+1;
        long sum=0, mod=1000000007;
        for(int i=st;i<=en;i++){
            sum+=(dp[i]+n-1);
            sum%=mod;
        }
        for(int i=n-1;i>=1;i--){
            st--;
            sum+=dp[st];
            sum-=dp[en];
            en--;
            sum+=mod;
            sum%=mod;
            dp[i]=(dp[i+1]+sum)%mod;
        }
        out.println(dp[1]);
        out.close();

    }
}
