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 template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author mehrdad
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int a=0,b;
        long time=0;
        for(int i=0;i<n;i++)
        {
            b=sc.nextInt();
            //if(i==0) time-=b;
            if(b>=a)
            {
                time+=b-a+2;//go up
            }
            else if(b<a)
            {
                time+=a-b+2;
            }
            a=b;
        }
        System.out.print(time-1);//minus 1 for first tree!
    }
}
