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
///*to calculate total number of times in eating all nuts of a tree*/
import java.util.*;
public class cf14
{
	public static void main(String args[])
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt(),i=0,time=0;
		int []a=new int[n];
		for(i=0;i<n;i++)
		{
			a[i]=scan.nextInt();
		}
		time=a[0]+1;
		for(i=1;i<n;i++)
		{
			if(a[i]<a[i-1])
			{
				//time=time+a[i-1]+a[i]+1;
				time=time+a[i-1]-a[i]+1+1;
				//System.out.println("time="+time);
			}
			else
			{
				time=time+1+a[i]-a[i-1]+1;
				//System.out.println("time="+time);
			}	
		}
		System.out.println(time);
	}
}