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.*;
import java.util.*;

public class RoadsideTrees {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int arr[] = new int[n + 1];
		int sum = 0;
		arr[0] = 0;
		for (int i = 1; i <= n; i++)
			arr[i] = sc.nextInt();
		for (int i = 1; i < n; i++) {
			if (arr[i] > arr[i - 1])
				sum += (arr[i] - arr[i - 1]);
			if (arr[i] > arr[i + 1])
				sum += (arr[i] - arr[i + 1]);
		}
		if (arr[n] > arr[n - 1])
			sum += (arr[n] - arr[n - 1]) + 1;
		else
			sum += 1;
		System.out.println(sum + (n - 1) * 2);
	}
}
