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.*;
public class hero{

 public static void main(String[] args) {
  Scanner no=new Scanner(System.in);
  int n=no.nextInt();
  int sum=0;
  int arr[]=new int[n];
  for(int i=0;i<n;i++){
    arr[i]=no.nextInt();
  }
  sum=arr[0]+n+(n-1);
  for(int j=1;j<n;j++){
    if(arr[j]>arr[j-1]){
      sum=sum+arr[j]-arr[j-1];
    }
    else if(arr[j]<arr[j-1]){
      sum=sum+arr[j-1]-arr[j];
    }
  }
  System.out.println(sum);
 }
}
  