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 oo = 1e9;
long long nmod(long long a) { return (((a % oo) + oo) % oo); }
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dxk[] = {1, 1, 2, 2, -1, -1, -2, -2};
int dyk[] = {2, -2, 1, -1, 2, -2, 1, -1};
int r, c;
bool val(int r1, int c1) { return (r1 > -1 && r1 < r && c1 > -1 && c1 < c); }
int ar[100001];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ar[i]);
  }
  int sm = 0;
  int here = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      here = ar[i];
      sm += ar[i] + 1;
      continue;
    }
    if (here <= ar[i]) {
      sm += ar[i] + 2 - here;
    } else if (here > ar[i]) {
      sm += here - ar[i] + 2;
    }
    here = ar[i];
  }
  printf("%d", sm);
  return 0;
}
