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.io.*;
import java.util.*;
public final class roadside_trees
{
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	static long min=0;
	
	static void solve(long[] a,int n)
	{
		min=a[0];
		for(int i=0;i<n;i++)
		{
			min++;
			if(i<n-1)
			{
				min++;
				if(a[i]>a[i+1])
				{
					min+=a[i]-a[i+1];
				}
				else if(a[i]<a[i+1])
				{
					min+=a[i+1]-a[i];
				}
				else
				{
					continue;
				}
			}
			else
			{
				continue;
			}
		}
		out.println(min);
	}
	
	public static void main(String args[]) throws Exception
	{
		int n=sc.nextInt();
		long[] a=new long[n];
		for(int i=0;i<n;i++)
		{
			a[i]=sc.nextLong();
		}
		solve(a,n);
		out.close();
	}
}