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 MAX 200000

int main() {
	int N, A[MAX];
	long long cnt = 0;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &A[i]);
	}

	for (int i = 0; i < N; i++) {
		if (i == 0) {
			if (A[0] != 0) {
				cnt = -1;
				break;
			}
		}
		else if (A[i] > A[i - 1] + 1) {
			cnt = -1;
			break;
		}
		else if (A[i] == A[i - 1] + 1) {
			cnt++;
		}
		else if (A[i] <= A[i - 1]) {
			cnt += A[i];
		}
	}

	printf("%lld\n", cnt);


	return 0;
}