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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;


class SequenceGrowingEasy{
    static void Main(string[] args){
        var n = int.Parse(Console.ReadLine());
        var aArray = new int[n];
        for(int i=0; i<n; i++){
            aArray[i] = int.Parse(Console.ReadLine());
        }

        if(aArray[0] != 0){
            Console.WriteLine(-1);
            return;
        }

        var sum = 0;
        var count = 0;
        for(int i=1; i<n; i++){
            sum = sum + aArray[i];
            if(aArray[i] != 0 && !(aArray[i-1] == aArray[i] - 1 || aArray[i] <= aArray[i-1])){
                Console.WriteLine("{0}", -1);
                return;
            }
            if(aArray[i] - aArray[i-1] == 1 && aArray[i-1] != 0){
                count = count + 1;
            }
        }

        Console.WriteLine(sum-count);
    }
}