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>
typedef long long ll;

ll n;
ll a[2 * 100005];

int main()
{
	ll ans = 0;

	scanf("%lld", &n);
	for (ll i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
	}

	a[0] = -1;
	for (ll i = n; i >= 0; i--) {
		if (a[i] - a[i - 1] > 1) {
			printf("-1\n");
			return 0;
		}
		else {
			if (a[i] > 0) {
				if (a[i] - a[i - 1] == 1) {
					ans++;
				} else {
					ans = ans + a[i];
				}
			}
		}
	}
	
	printf("%lld\n", ans);

	return 0;
}