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
class Task:
	h = []
	answer = 0
	
	def getData(self):
		h_size = int(input())
		self.h = [int(input()) for i in range(h_size)]
	
	def solve(self):
		current_level = 0
		for i in range (0, len(self.h)):
			self.answer += self.h[i] - current_level + 1
			if i == len(self.h) - 1:
				return
			if self.h[i + 1] < self.h[i]:
				self.answer += self.h[i] - self.h[i + 1]
				current_level = self.h[i + 1]
			else:
				current_level = self.h[i]
			self.answer += 1
	
	def printAnswer(self):
		print(self.answer)


task = Task();
task.getData();
task.solve();
task.printAnswer();
