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 TreesAlongRoadSolver {

    private int n;

    private int[] h;

    public static void main(String[] args) {
        TreesAlongRoadSolver solver = new TreesAlongRoadSolver();

        solver.readData();
        long result = solver.solve();

        System.out.println(String.format("%d", result));
    }

    private long solve() {
        long result = -1;
        int position = 0;

        for (int i = 0; i < n; i++) {
            result += 2 + Math.abs(h[i] - position);
            position = h[i];
        }

        return result;
    }

    private void readData() {
        Scanner scanner = new Scanner(System.in);

        n = scanner.nextInt();

        h = new int[n];
        for (int i = 0; i < n; i++) {
            h[i] = scanner.nextInt();
        }
    }
}

