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.*;
import java.lang.*;
import java.io.*;

public class S
{
	public static long M = 1000000007;
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int h[] = new int[n];
		for(int i=0; i<n; i++){
			h[i] = sc.nextInt();
		}
		if(n==1){
			System.out.println("2");
			return;
		}
		long sum = h[0] ;
		for(int i=0; i<n-1; i++){
			sum = sum + 1 + (h[i] - Math.min(h[i], h[i+1])) + 1 + (h[i+1]- Math.min(h[i], h[i+1]));
			//System.out.println(sum+" "+Math.min(h[i], h[i+1])+" "+h[i]+" "+h[i+1]);
		}
		System.out.println(sum+1);
	}
}