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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import javafx.scene.shape.Line;
/*
 * Problem B in O(1)
 * 
 * 
 * 
 * 
 * 
*/
public class Main {
	public static void main(String []args){
		Scanner sc= new Scanner(System.in) ;
       
        int n = sc.nextInt();
        int ans =0 ;int arr[] = new int[n]; 
        for(int i=0 ; i <n; i++)arr[i]=sc.nextInt();
        int prev= arr[0];
        ans = prev +1; 
		for(int i =1 ; i<n ; i++)
		{
			int tmp =arr[i];
			if(prev<=tmp) ans+=(tmp-prev)+1+1 ;
			else
			{
				ans+=prev-tmp+2;
			}
		
		prev= tmp ;
		}
		System.out.println(ans);
		
		

		
	
		
	}

	
	
	
	
}
	
	
	
	
	



	 
	
	
	
	

	





