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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int ans = 0;
		boolean flg = false;
		int[] a = new int[n];
		for(int i = 0; i < n; i++) {
			a[i] = scanner.nextInt();
		}
		if (a[0] != 0) flg = true;
		for(int r = n - 1; r > 0; r--) {
			if(a[r] - a[r-1] > 1) {
				flg = true;
				break;
			}
			if(a[r] != 0) {
				if(a[r] <= a[r-1]) {
					ans += a[r];
				}else {
					ans++;
				}
			}
			
		}
		if(flg) {
			System.out.println(-1);
		}else {
			System.out.println(ans);
		}
	}

}