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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
template <typename T>
string itos(T i) {
  stringstream second;
  second << i;
  return second.str();
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  ll res = 0;
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  ll ini = 0;
  for (int i = 0; i < n; i++) {
    res += arr[i] - ini;
    if (i == n - 1) break;
    if (arr[i + 1] >= arr[i]) {
      res++;
      ini = arr[i];
    } else if (arr[i + 1] < arr[i]) {
      res += arr[i] - arr[i + 1];
      res++;
      ini = arr[i + 1];
    }
  }
  cout << res + n;
}
