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<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn=200002;
long long a[maxn];
int main()
{
	int n,flag=0;
	long long ans=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		if(a[1]!=0||(a[i]-a[i-1]>1))
			flag=1;
	}
	if(flag)
	{
		printf("-1");
		return 0;
	}
	for(int i=1;i<=n;i++)
	{
		if(!a[i])				continue;
		else if(a[i]==a[i-1]+1)	ans++;
		else					ans+=a[i];
	}
	printf("%d",ans);
	return 0;
}