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.*;
import java.io.*;
public class roadside{
	public static void main(String arg[])
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int h[]=new int[n];
		for(int i=0;i<n;i++)
		{
			h[i]=sc.nextInt();
		}
		int s=0;
		for(int i=0;i<n;i++)
		{ 
	         if(s!=0)
			 {
				 if(h[i-1]<h[i])
				 {
					// s=s+(h[i]-h[i-1])+1;
                       s=s+1+(h[i]-h[i-1])+1;
				 }
				 else if(h[i-1]>h[i])
				 {
					 s=s+(h[i-1]-h[i])+1+1;
				 }
				 else{
					 s=s+1+1;
					 
				 }
			 }
				 
				 
				 
				 
				 
				   if(s==0)
			    {
				    s=s+h[i]+1;
					/*if(h[i+1]<h[i])
					{
						s=s+(h[i]-h[i+1])+1;
					}
					else if(h[i+1]>h[i])
					{
						s=s+1;
					}
					else{
						s=s+1;
					}*/
			     }
		}
				 
			
		System.out.println(s);
	}
}