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 JavaApplication16{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int a[]=new int[n+2];
for(int k=1;k<=n;k++){
a[k]=sc.nextInt();
}
int q=0;
for(int k=1;k<n;k++){
if(k==1){
   q=a[k]+1;
   if(a[k]>a[k+1]){
      q=q+a[k]-a[k+1]+2;   
     }
     if(a[k]<=a[k+1]){
         
         q=q+a[k+1]-a[k]+2;
     
}
}


  else if(k<n&&k!=1){
     if(a[k]>a[k+1]){
      q=q+a[k]-a[k+1]+2;   
     }
     if(a[k]<=a[k+1]){

         q=q+a[k+1]-a[k]+2;
   }   
  }}
if(n==1){
    q=q+a[1]+1;
}
System.out.println(q);


}}