Problem:
Score : 700 points

Problem Statement

There is a sequence X of length N, where every element is initially 0. Let X_i denote the i-th element of X.

You are given a sequence A of length N. The i-th element of A is A_i. Determine if we can make X equal to A by repeating the operation below. If we can, find the minimum number of operations required.

Choose an integer i such that 1\leq i\leq N-1. Replace the value of X_{i+1} with the value of X_i plus 1.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 10^9(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

If we can make X equal to A by repeating the operation, print the minimum number of operations required. If we cannot, print -1.

Sample Input 1

4
0
1
1
2

Sample Output 1

3

We can make X equal to A as follows:

Choose i=2. X becomes (0,0,1,0).

Choose i=1. X becomes (0,1,1,0).

Choose i=3. X becomes (0,1,1,2).

Sample Input 2

3
1
2
1

Sample Output 2

-1

Sample Input 3

9
0
1
1
0
1
2
2
1
2

Sample Output 3

8

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <iostream>

using namespace std;

long a[200001];
long long tb[200001][2];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];

  if (a[1] != 0) {
    cout << -1 << endl;
    return 0;
  }

  tb[0][0] = 0;
  tb[0][1] = 0;
  for (int i = 2; i <= n; i++) {
    tb[i][0] = min(tb[i - 1][0], tb[i - 1][1]) + a[i];
    if (a[i] > a[i - 1]) {
      if (a[i - 1] == 0) {
        tb[i][1] = min(tb[i - 1][0], tb[i - 1][1]) + a[i];
      } else {
        tb[i][1] = min(tb[i - 1][0], tb[i - 1][1]) + 1;
      }
    } else {
      tb[i][1] = tb[i][0];
    }
  }
  cout << min(tb[n][0], tb[n][1]) << endl;

  return 0;
}