/*
# 问题分析
1. 题设要求模拟电梯的升降，并计算完成所有楼层停靠要求所需要的总时间。
2. 由于电梯停考规则和停靠时间题设已经明确给出, 所以这是一道只需要我们按照规则翻译成代码的模拟题。
 
# 完整描述步骤
1. 获取输入: 楼层停靠请求数目
2. 初始化记录器:
    - 当前楼层 = 0
    - 总计耗时 = 0
3. 依次读入楼层停靠请求:
    - 如果要前往的楼层层次比当前楼层小:
        - 总计耗时 += (当前楼层 - 目标楼层) * 4 + 5;
    - 如果要前往的楼层层次比当前楼层大:
        - 总计耗时 += (目标楼层 - 当前楼层) * 6 + 5;
    - 如果要前往的就是当前楼层:
        - 总计耗时 += 5;
    - 设置读入的目标楼层为当前楼层
 
5. 输出总计耗时
 
# 伪代码描述
1. get input: request_amount
2. init recorders:
    - current_layer = 0
    - total_time_cost = 0
3. for each request:
    - get input: target_layer
    - if target_layer < current_layer:
        - total_time_cost += (current_layer - target_layer) * 4;
    - elif target_layer > current_layer:
        - total_time_cost += (target_layer - current_layer) * 6;
    - total_time_cost += 5;
    - current_layer = target_layer;
4. print(total_time_cost)
 
*/
 
# include<iostream>
using namespace std;
 
int main(){
    int request_amount;
    cin >> request_amount;
    int current_layer = 0;
    int total_time_cost = 0;
    for (int i = 0; i < request_amount; i++){
        int target_layer;
        cin >> target_layer;
        if (target_layer > current_layer){
            total_time_cost += (target_layer - current_layer) * 6;
        } else if (target_layer < current_layer) {
            total_time_cost += (current_layer - target_layer) * 4;
        }
        total_time_cost += 5;
        current_layer = target_layer;
    }
    
    cout << total_time_cost;
    return 0;
}