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 Max {

    public static void main(String[] args) {
    
        
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt();
        int[] heights = new int[n];
        for(int i = 0; i < n; i++)
            heights[i] = sc.nextInt();
        int time = 0;int h = 0;
        for(int i = 0; i < n; i++)
        {
            time += heights[i] - h;
            h = heights[i];
            time++;

            if(i<n-1)
            {
                time++;
                if(h>heights[i+1])
                {
                    time += h - heights[i+1]; h = heights[i+1];
                }
            }
            
            
            
        }
        System.out.println(time);
            
    }
    

}
