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;

public class Main {
	static Scanner s;

	

	public static void main(String[] args) {
		{int n;
		int steps;
		 s= new Scanner(System.in);
		n=s.nextInt();
		int a[]=new int[n];
			for(int i=0;i<n;i++)
			{a[i]=s.nextInt();}
			steps=a[0];
			steps++;
			for(int i=1;i<n;i++)
			{if(a[i-1]<=a[i])
			 {
			steps++;	
			 steps=steps+(a[i]-a[i-1]);			 
			 steps++;
			 }
			
			else
			{ steps=steps+(a[i-1]-a[i]);
			steps++;
			steps++;
			
			}
			}
System.out.println(steps);
			
		}
	}

}
