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 roadsideTrees {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a[] = new int[n];
            for (int i = 0; i < a.length; i++) 
            {
                a[i] = sc.nextInt();    
            }
        int up=0,down=0,eat=n,jump=0;
        up = a[0];
        for(int i =0;i<n-1;i++)
            {
                if(a[i]<a[i+1])
                {
                    jump++;
                    int count = a[i+1]-a[i];
                    up = up+count;
                }
                else
                {
                    int count = a[i]-a[i+1];
                    down = count+down;
                    jump++;
                }
            }
            System.out.println(jump+up+down+eat);
            sc.close();
    }
}