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
#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
template <typename T>
void printV(vector<T> v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] % 1000000007 << " ";
  }
  cout << endl;
}
template <typename T>
void printA(T arr[], long long n) {
  for (int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
template <typename T>
void printPA(T* arr, long long n) {
  for (int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
bool isSorted(int* arr, int n) {
  for (int i = 0; i < n - 1; i++) {
    if (arr[i] > arr[i + 1]) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int time = 0;
  int i = 0;
  int pos = 0;
  while (i < n - 1) {
    time += (arr[i] - pos) + 1;
    pos = arr[i];
    if (pos > arr[i + 1]) {
      time += (pos - arr[i + 1]) + 1;
      pos = arr[i + 1];
    } else {
      time++;
    }
    i++;
  }
  time += arr[n - 1] - pos + 1;
  cout << time;
  return 0;
}
