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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
class Program
{
	static void Main(string[] args)
	{
		long N = long.Parse(Console.ReadLine());
		long [] P = new long[N];
		long count = 0;
		P[0] = long.Parse(Console.ReadLine());
		if(P[0] > 0){
			count = -1;
		} else {
			for(var i=1;i<N;i++){
				P[i] = long.Parse(Console.ReadLine());
				if(P[i] - P[i-1] == 1){
					count += 1;
				} else if(P[i] - P[i-1] >= 2){
					count += P[i] / (P[i-1] + 1) + P[i] % (P[i-1] + 1);
				} else if(i>=2 && P[i-1] - P[i-2] >= 2){
					if(P[i] - 1 <= P[i-1] % (P[i-2] + 1)){
						count += 1;
					} else if ((P[i] - 1) % P[i-1] == P[i-2] + 1){
						count += 1;
					} else {
						count += P[i];
					}
				} else {
					count += P[i];
				}
			}
		}
		Console.WriteLine(count);
	}
}