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.*;
public class B14
{

	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in); 
		
		int n=sc.nextInt();
		int arr[] = new int[n];
		for( int i=0; i<n; i++ )
			arr[i]=sc.nextInt();
		
		int ans=0;
		
		if( n==1 )
		{
			ans+=arr[0];
		}
	
		for( int i=0; i<=n-2; i++ )
		{
			if( i==0 )
			{
				ans+=arr[0];
			}
			
			if( arr[i]>arr[i+1] )
			{
				ans+= arr[i]-arr[i+1];
			}
			
			if( arr[i]<arr[i+1] )
			{
				ans+= arr[i+1]-arr[i];
			}
			
			
			ans++;
		}
		
		ans+= n;
		
		System.out.println(ans);
			
	}

}

