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 hell = 1000000007;
long long int bina(long long int a, long long int b) {
  long long int res = a;
  long long int c = a;
  long long int ans = 1;
  while (b) {
    ans = ans % hell;
    if (b & 1) {
      ans *= res;
    }
    res = (res * res) % hell;
    b >>= 1;
  }
  return ans % hell;
}
void solve() {
  long long int a = 1, b = 0, c = 0, t = 0, ans = -1, k = 1, ans2 = 0, c1 = 0,
                c2 = 0, n = 0, m = 0, x = 2231131199;
  cin >> a;
  vector<long long int> v(a);
  for (int i = 0; i < a; ++i) cin >> v[i];
  ;
  ans = 2 * a - 1;
  for (int i = 0; i < a; ++i) {
    ans += abs(v[i] - c);
    c = v[i];
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t z = clock();
  long long int qc = 1;
  for (long long int i = 1; i <= qc; i++) solve();
  fprintf(stderr, "Total Time: % .4Lf\n",
          (long double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
}
