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];
      int[] b = new int[N]; 
      int count =0;
      for(int i=0;i<N;i++){
       a[N-i-1]=sc.nextInt();
       b[i] = 0;
      }
      for(int j=0;j<N;j++){
                  if(a[j]==b[j]){continue;}
                  for(int k=0;k<a[j];k++){
                           b[j+k]=a[j]-k;
                           count++;
                  }
      }
      for(int q=1;q<N;q++){
            if(a[q]>=2&&a[q]-a[q-1]>=2){
                     count=-1;
            }else if(a[N]!=0){count=-1;}
      }
      System.out.println(count);
   }
}