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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
using namespace std;
int main()
{
	int N;
	cin >> N;
	int A[N];
	for (int i = 0; i < N; i++) {
		if (scanf("%d", &A[i]) == EOF) {
			cout << "ERROR" << endl;
			return 0;
		}
	}
	if (A[0] > 0) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 1; i < N; i++) {
		if (A[i] - A[i - 1] > 1) {
			cout << -1 << endl;
			return 0;
		}
	}
	int ans = 0;
	for (int i = 1; i < N; i++) {
		if (A[i - 1] + 1 == A[i]) {
			ans ++;
		}
		else {
			ans += A[i];
		}
	}
	cout << ans << endl;
	return 0;
}