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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
ll n,a[N];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	ll res=0,pre=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]>=i)
		{
			res=-1;
			break;
		}
		if(a[i]>pre)
		{
			if(a[i]-pre>1)
			{
				res=-1;
				break;
			}
			res++;
		}
		else
			res+=a[i];
	//	cout<<i<<' '<<res<<'\n'; 
		pre=a[i];		
	}
	cout<<res<<'\n';
	return 0;
}