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<cstdlib>
#include<iostream>
#include<queue>
#include<string>
#include<map>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std ;
const int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int n;
long long a[200000+5];
int main()
{
    while(~scanf("%d",&n))
    {
        bool flag=1;
        long long sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
        }
        int x=-1;
        for(int i=0;i<n;i++)
        {
            x++;
            if(!flag) break;
            if(a[i]>x) {flag=0;break;}
            if(a[i])
            {
               if(a[i-1]>=a[i]) sum+=a[i];
               else if(a[i]-a[i-1]==1) sum++;
               else flag=0;
            }
            else
            {
                x=0;
            }
        }
        printf("%lld\n",flag?sum:-1);

    }

    return 0;
}