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<bits/stdc++.h>
using namespace std;
#define int long long
int a[200005];
signed main(){
	int n;
	cin >> n;
	bool check;
	for(int i = 1 ; i <= n ; i++)
	{
		cin >> a[i];
		if(i == 1)
		{
			if(a[i] != 0)
			{
				check = 1;
			}
		}
		else
		{
			if(a[i] > a[i - 1] + 1)
			{
				check = 1;
			}
		}
	}
	a[n + 1] = 1e9 + 7;
	if(check == 1) { cout << -1; return 0; }
	int i = 1;
	int ans = 0;
	while(i <= n)
	{
		if(a[i] == 0)
		{
			i++;
			continue;
		}
		int t = a[i];	
		while(a[i + 1] == a[i] + 1)
		{
			i++;
			t++;
		}
		ans += t;
		i++;
	}
	cout << ans;
}