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: Accepted)

#include<stdio.h>
int main(){
  long a[200000],cnt=0;
  long i,n;
  scanf("%ld",&n);
  for(i=0;i<n;i++){
    scanf("%ld",&a[i]);
  }
  if(a[n-1]>n-1){
    printf("-1\n");
    return 0;
  }
  else
    cnt=a[n-1];
  for(i=n-2;i>=0;i--){
    if(a[i+1]>a[i]+1||a[i]>i){
      printf("-1\n");
      return 0;
    }
    else if(a[i+1]!=a[i]+1)
      cnt+=a[i];
  }
  printf("%ld\n",cnt);
}