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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

uint32_t A[200000];

static int64_t calcMinAddCount(const uint32_t N) {
	if (A[0] != 0) {
		return INT64_C(-1);
	}

	int64_t minAddCount = 0;

	for (uint32_t idx = 1; idx < N; idx++) {
		const int64_t diff = (int64_t)(A[idx]) - (int64_t)(A[idx - 1]);

		if (diff > 1) {
			return INT64_C(-1);
		}
		else if (diff == 0) {
			minAddCount += (int64_t)(A[idx]);
		}
		else if (diff == 1){
			minAddCount++;
		}
		else {
			minAddCount += (int64_t)(A[idx]);
		}
	}

	return minAddCount;
}

int main(void) {
	uint32_t N;

	scanf("%"PRIu32"\n", &N);
	for (uint32_t idx = 0; idx < N - 1; idx++) {
		scanf("%"PRIu32"\n", &(A[idx]));
	}
	scanf("%"PRIu32, &(A[N - 1]));

	const int64_t minAddCount = calcMinAddCount(N);

	printf("%"PRId64, minAddCount);

	return 0;
}