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>

#define N 200000

long a[N+1];

int main(void){
    long n;
    scanf("%ld", &n);
    a[0] = -1;
    for(long i=1; i<=n; i++){
        scanf("%ld", a+i);
    }
    long count=-1;
    for(long i=1; i<=n; i++){
        long d = a[i]-a[i-1];
        if(d>=2){
            puts("-1");
            return 0;
        }else if(d==1){
            count++;
        }else{
            count += a[i];
        }
    }
    printf("%ld\n", count);
    return 0;
}