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 CodeforcesRound162B {

    /**
     * @param args
     */
    public static void main(String[] args) {
    // TODO Auto-generated method stub
    
    Scanner kde=new  Scanner(System.in);
    
    int n=kde.nextInt();
    int[] h= new int[n];
    for(int i=0; i<n; i++ )
    {
       h[i]=kde.nextInt(); 
    }
    int sum=0;
       sum=h[0]+1;
    int visota=h[0];  
       
    for(int i=1; i<n; i++ )
    {
     if(visota<=h[i])
     {
         sum=sum+h[i]-visota+2;
         visota=h[i];
     }
     else 
     {
         sum=sum+(visota-h[i])+2;
         visota=h[i];
     }    
        
    }
    
    
     System.out.println(sum);   
        
        
        
    

    }

}
