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 Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner keyboard = new Scanner(System.in);
		int arboles = keyboard.nextInt();
		int [] alturas = new int [arboles];
		
		for(int i = 0; i < arboles; i++) {
			alturas[i] = keyboard.nextInt();
		}
		
		//Para contar diferencia entre arboles
		int movimientos = 0;
		for(int i = 0; i < (alturas.length-1); i++) {
			if(alturas[i] > alturas[i+1]) {
				movimientos = movimientos + (alturas[i] - alturas[i+1]);
			} else if(alturas[i] < alturas[i+1]) {
				movimientos = movimientos + (alturas[i+1] - alturas[i]);
			}
		}
		
		//Suma de todos los movimientos
		System.out.print(movimientos + (arboles-1) + arboles + alturas[0]);
		
		
		

	}

}
