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: Runtime Error)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        boolean able = true;
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            if(i < a[i]) able = false;
            if(0 < i && a[i-1] < a[i] && 1 < a[i] - a[i-1]) able = false;
        }
        if(!able){
            System.out.println(-1);
        }else{
            long ans = 0L;
            int[] b = new int[n];
            Arrays.fill(b, -1);
            for (int i = n-1; i > 0; i++) {
                if(b[i] != a[i]){
                    ans += a[i];
                }
                b[i-1] = a[i] - 1;
            }
            System.out.println(ans);

        }
        sc.close();

    }

}