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<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;

const int N=2e5+1e3+7;

int n,a[N];

ll ans;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	if(a[1]!=0)
	{
		puts("-1");
		return 0;
	}
	for(int i=2;i<=n;i++)
	{
		if(a[i]>a[i-1]+1)
		{
			puts("-1");
			return 0;
		}
		else if(a[i]==a[i-1]+1)
			continue;
		else
			ans+=a[i-1];
	}
	ans+=a[n];
	printf("%lld",ans);
}