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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.TreeSet;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner scanner = new Scanner(System.in);

		int N = scanner.nextInt();

		List<Long> alist = new ArrayList<>();
		List<Long> xlist = new ArrayList<>();

		for (int i = 0; i < N; i++) {
			alist.add(scanner.nextLong());
		}

		if (alist.get(0) != 0) {
			System.out.println("-1");
			return;
		}

		long count = 0;

		for(int i = 0; i < N-1; i++) {
			if (alist.get(i+1) == alist.get(i) + 1) {
				count = count + 1;
			} else if (alist.get(i+1) == alist.get(i)) {
				count = count + alist.get(i+1) ;
			} else if (alist.get(i+1) == alist.get(i) - 1) {
				count = count + alist.get(i+1);
			}
		}

		System.out.println(count);



	}

}