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 *h*<sub class="lower-index">*i*</sub>. 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* &gt; *h*<sub class="lower-index">*i* + 1</sub>. 
Compute the minimal time (in seconds) required to eat all nuts.
The first line contains an integer *n* (1  ≤  *n* ≤ 10<sup class="upper-index">5</sup>) — the number of trees.

Next *n* lines contains the height of trees: *i*-th line contains an integer *h*<sub class="lower-index">*i*</sub> (1 ≤ *h*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">4</sup>) — the height of the tree with the number *i*.
import time,math
from sys import stdin,stdout
def I():
    return int(stdin.readline())
def In():
    return map(int,stdin.readline().split())
def L():
    return list(map(int,stdin.readline().split()))
def Si():
    return sorted(list(map(int,stdin.readline().split())))
def Si():
    return sorted(list(map(int,stdin.readline().split())),reverse=True)
def P(x):
    return stdout.write(str(x))
def Pi(x,y):
    return stdout.write(str(x)+" "+str(y))  
 
def solve():
    n=I()
    boo=I()
    ans=boo+1
    for i in range(n-1):
        t=I()
        ans+=2
        if(t!=boo):
            ans+=abs(boo-t)
    P(ans)              
 
if __name__ == "__main__" :
    solve()