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.Arrays;

/**
 *
 * @author Gitesh
 */
public class Main {

   
    public static void main(String[] args) throws IOException{
        
        InputStreamReader inp = new InputStreamReader(System.in);
    BufferedReader br = new BufferedReader(inp);
 int n=Integer.parseInt(br.readLine());
 int h[]=new int[n];
 for(int i=0;i<n;i++)
 {
 h[i]=Integer.parseInt(br.readLine());
 }
 
 int time=h[0]+1;
 for(int j=1;j<n;j++)
 {
     if(h[j-1]>h[j])
         time+=h[j-1]-h[j]+2;
     else
     { time+=h[j]-h[j-1]+2; }
     
 }
  System.out.println(time);
  

          
    }
   



}
