Squirrel Liss loves nuts. There are n trees (numbered 1 to n from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree i is hi. Liss wants to eat all nuts.

Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:

  * Walk up or down one unit on a tree. 
  * Eat a nut on the top of the current tree. 
  * Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height h of the tree i (1 ≤ i ≤ n - 1), she jumps to height h of the tree i + 1. This action can't be performed if h > hi + 1. 



Compute the minimal time (in seconds) required to eat all nuts.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of trees.

Next n lines contains the height of trees: i-th line contains an integer hi (1 ≤ hi ≤ 104) — the height of the tree with the number i.

Output

Print a single integer — the minimal time required to eat all nuts in seconds.

Examples

Input

2
1
2


Output

5


Input

5
2
1
2
1
1


Output

14

import java.util.Scanner;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author KHALED
 */
public class RoadsideTrees_SimplifiedEdition {
    public static void main(String[] args) {
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
//        int[]arr=new int[n];
//        for (int i = 0; i < n; i++) 
//            arr[i]=scan.nextInt();
//        if(n==1)
//        {
//            System.out.println(arr[0]+1);
//            return;
//        }
//        int sum1=0,sum2=0;
//        sum1+=arr[0]+1;
//        for (int i = 1; i < n; i++) 
//        {
//            if(arr[i]<arr[i-1])
//                sum1+=1;
//            else
//            {
//                sum1+=arr[i-1];
//                sum1+=arr[i];
//            }
//            sum1++;
//        }
//        sum2+=arr[n-1]+1;
//        for (int i = n-2; i >= 0; i--) 
//        {
//            if(arr[i]<arr[i+1])
//                sum2+=1;
//            else
//            {
//                sum2+=arr[i+1];
//                sum2+=arr[i];
//            }
//            sum2++;
//        }
//        System.out.println(Math.min(sum1, sum2));
        long ans=0;
        int prev=0;
        int cur=0;
        for (int i = 0; i < n; i++) {
            cur=scan.nextInt();
            if(cur>=prev){ans++;
            ans+=(cur-prev);
            ans++;
            }
           else{
           ans+=(prev-cur);
           ans+=2;
           
           }
           prev=cur;
       }
   ans--;
       System.out.println(ans);
    }
}
