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 mul(long long int a, long long int b) {
  return ((a % int(1e9 + 7)) * (b % int(1e9 + 7))) % int(1e9 + 7);
}
long long int add(long long int a, long long int b) {
  return ((a % int(1e9 + 7)) + (b % int(1e9 + 7))) % int(1e9 + 7);
}
long long int sub(long long int a, long long int b) {
  long long int ans = ((a % int(1e9 + 7)) - (b % int(1e9 + 7))) % int(1e9 + 7);
  ans %= int(1e9 + 7);
  ans = (ans + int(1e9 + 7)) % int(1e9 + 7);
  return ans;
}
long long int mpow(long long int a, long long int b) {
  long long int ans = 1;
  long long int po = a;
  while (b != 0) {
    if (b % 2) {
      ans = ((ans % int(1e9 + 7)) * (po % int(1e9 + 7))) % int(1e9 + 7);
    }
    po = ((po % int(1e9 + 7)) * (po % int(1e9 + 7))) % int(1e9 + 7);
    b /= 2;
  }
  return ans;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> h(n);
  for (int i = 0; i < n; i++) cin >> h[i];
  long long int ans = 1 + h[0];
  for (int i = 0; i < n - 1; i++) {
    ans += (2 + abs(h[i] - h[i + 1]));
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  solve();
}
