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
// https://codeforces.com/problemset/problem/265/B

import java.util.Scanner;

public class Solution {

	public static void main(String[] args) {
		Scanner read = new Scanner(System.in);
		
		int n = read.nextInt();
		
		int[] h = new int[n];
		for (int i = 0; i < n; ++i) {
			h[i] = read.nextInt();
		}
		
		int currentHeight = 0;
		int minNumberOfSeconds = 0;
		
		for (int i = 0; i < n; ++i) {
			if (currentHeight <= h[i]) {
				minNumberOfSeconds += (h[i] - currentHeight); // climb up until nut is reached
			}
			else {
				minNumberOfSeconds += (currentHeight - h[i]); // climb down until nut is reached
			}
			minNumberOfSeconds++; // eat nut
			currentHeight = h[i]; // you are now at the top of the tree
			if (i+1 < n) { // if there is a next tree
				if (currentHeight > h[i+1]) {
					minNumberOfSeconds += (currentHeight - h[i+1]);
					currentHeight = h[i+1];
				}
				minNumberOfSeconds++; // jump to next tree
			}
		}
		
		System.out.println(minNumberOfSeconds);
		
		// Close scanner
		read.close();
	}

}