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 ConsoleApplication {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int[] arr;
        arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
        System.out.println(getMinimumNoOfMoves(arr));
    }//End main funciton.

    public static int getMinimumNoOfMoves(int[] arr) {

        int count = 0;
        int action = 2;

        for (int i = 0; i < arr.length; i++) {

            if (i == 0) {
                count += arr[i] + 1;
            } else {
                if(arr[i]>=arr[i-1]){
                    count += (arr[i]-arr[i-1]) + action;
                }else{
                    count += (arr[i - 1]-arr[i]) + action;
                }
            }
        }

        return count;
    }

}//End ConsoleApplication Class.
