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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int N, i;
int A[200000];
int ans;

int main(){
  scanf("%d", &N);
  for(i=0; i<N; i++){
    scanf("%d", A+i);
  }
  if(A[0] != 0){
    printf("-1\n");
    return 0;
  }
  for(i=0; i<N-1; i++){
    if(A[i]+1 < A[i+1]){
      printf("-1\n");
      return 0;
    }
  }
  for(i=1; i<N; i++){
    if(A[i-1]+1 == A[i]){
      ans++;
    }else{
      ans += A[i];
    }
  }
  printf("%d\n", ans);
}