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 mpow(long long a, long long b, long long c = 1000000007) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % c) * (a % c)) % c;
    a = ((a % c) * (a % c)) % c;
    b >>= 1;
  }
  return res;
}
unordered_set<long long> prime;
void seive() {
  int n = 1e6;
  vector<char> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
      for (int j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  for (long long i = 0; i < n; i++) {
    if (is_prime[i]) prime.insert(i);
  }
}
void solve() {
  int n;
  cin >> n;
  int ar[n];
  for (long long i = 0; i < n; i++) {
    cin >> ar[i];
  }
  int result = ar[0] + 1;
  for (long long i = 0; i < n - 1; i++) {
    result += abs(ar[i] - ar[i + 1]);
    result += 2;
  }
  cout << result << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int loop = 1;
  while (loop--) {
    solve();
  }
  return 0;
}
