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 nuts
{
	public static void main(String args[])
	{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		int last=0,time=0,diff=0;
		int [] tree = new int[t];
		for(int j =0;j<t;j++)
		{
			tree[j]= sc.nextInt();
		}
		time = tree[0];
		for(int i =0;i<t-1;i++)
		{
			diff= tree[i+1]-tree[i];
			if(diff < 0)
			{
				diff = diff*(-1);

			}
			time =time+diff;
		}
		time = time +t +t-1;
		System.out.println(time);
	}
}