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;
long long int mod = 1000000007;
void dispvec(set<long long int> arr) {
  set<long long int>::iterator itr = arr.begin();
  while (itr != arr.end()) {
    cout << *itr << " ";
    itr++;
  }
  cout << endl;
}
struct CustomCompare {
  bool operator()(const pair<long long int, long long int> &i,
                  const pair<long long int, long long int> &j) {
    if (i.first == j.first) return i.first < j.first;
    return i.second < j.second;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<long long int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  long long int sum = arr[0] + 1;
  int curr = arr[0];
  for (int i = 1; i < n; i++) {
    if (arr[i] < arr[i - 1]) {
      sum += arr[i - 1] - arr[i];
    } else if (arr[i] > arr[i - 1]) {
      sum += arr[i] - arr[i - 1];
    }
    sum += 2;
  }
  cout << sum << endl;
}
int main() {
  int t = 1;
  int i = 1;
  while (i <= t) {
    solve();
    i++;
  }
  return 0;
}
