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: Runtime Error)

#include <stdio.h>

#define MAX 2*10000

int main(){
  int n;
  int a[MAX];
  int count = -1;
  int sa, t;

  scanf("%d\n", &n);
  for(int i = 0; i < n; i++){
    scanf("%d", &a[i]);
  }

  for(int i = 1; i < n; i++){
    t = i;
    sa = a[t--] - a[t];
    // printf("%d" ,sa);
  }

  if(a[0] == 0 && abs(sa) <= 1){
    count = 0;
    for(int i = 1; i <= n; i++){
      int b,c;
      b = i, c = i+1;

      if(a[i] != 0){
        // printf("%d ", a[i]);
        count++;
      }

      if(a[i] >= 2 && a[b] == a[c]){
        printf("%d%d", i, a[i]);
        count++;

        if(a[b] < a[c]){
          count++;
        }

      }

    }
  }

  printf("%d", count);
  // for(int i = 0; i < n; i++){
  //   printf("%d ", a[i]);
  // }

  return 0;
}