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.util.*;

public class Main {

  public static void main(String args[]) {
    (new Main()).solve();
  }

  void solve() {

    Scanner cin = new Scanner(System.in);

    int N = cin.nextInt();
    int array[] = new int[N];
    
    for(int i = 0; i < N; ++i ) {
    	array[i] = cin.nextInt();
    }
    
    int ans = 0;
    boolean add = true;
    for( int i = N - 1; i >= 0; --i ) {
    	if( array[i] == 0 ) {
    		add = true;
    		continue;
		}
    	if( add ) {
    		ans += array[i];
    	}
    	// failed...;
    	if( i == 0 ) {
    		ans = -1;
    		break;
    	}
    	if( array[i - 1] < array[i] - 1 ) {
    		ans = -1;
    		break;
    	}
    	if( array[i - 1] == array[i] - 1 ) {
    		add = false;
    	}
    	else {
    		add = true;
    	}
    }

	System.out.println(ans);
    
    cin.close();

  }

}