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.io.PrintWriter;
import java.util.Scanner;

/**
 * @author pvasilyev
 * @since 21 Oct 2016
 */
public class B265 {

    public static void main(String[] args) {
        final Scanner reader = new Scanner(System.in);
        final PrintWriter writer = new PrintWriter(System.out);

        solve(reader, writer);

        writer.close();
        reader.close();
    }

    private static void solve(final Scanner reader, final PrintWriter writer) {
        final int n = reader.nextInt();
        final int[] h = new int[n];
        for (int i = 0; i < n; ++i) {
            h[i] = reader.nextInt();
        }
        int sum = h[0] + 1;
        for (int i = 1; i < n; ++i) {
            if (h[i] < h[i-1]) {
                sum += (h[i-1] - h[i]) + 2;
            } else {
                sum += 1 + (h[i] - h[i-1]) + 1;
            }
        }
        writer.println(sum);
    }
}
