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;
struct pt {
  long long x, y;
};
long double dist(pt a, pt b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
int cmpr(pt a, pt b) {
  if (a.x == b.x)
    return a.y < b.y;
  else
    return a.x < b.x;
}
long double chek(pt a, pt b, pt c) {
  return (a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y)) / 2;
}
long long a[1111111];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  if (n == 1) return cout << a[0] + 1, 0;
  long long ans = 0;
  long long now = 0;
  for (long long i = 0; i < n - 1; i++) {
    if (now <= a[i])
      ans += a[i] - now;
    else {
      ans += a[i - 1] - a[i];
    }
    now = a[i];
    ans++;
    ans++;
  }
  if (now <= a[n - 1])
    ans += a[n - 1] - now;
  else {
    ans += a[n - 2] - a[n - 1];
  }
  ans++;
  cout << ans;
  return 0;
}
