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 <iostream>
#include <set>
#include <vector>

using namespace std;


int main(){
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    bool flag=true;
    int count=0;
    for(int i=0;i<n;i++){
        cin >> a[i];
        if(i==0){
            b[i]=0;
            if(a[i]!=0){
                flag=false;
            }
        }else{
            if (a[i]-a[i-1]>1)flag=false;
            b[i]=b[i-1]+((a[i]-a[i-1]==1)? 0 : 1);
            count+=a[i-1]*((a[i]-a[i-1]==1)? 0 : 1);
            if (i-a[i]<0 /*|| b[i]-b[i-a[i]]>1*/){//ほんまか？
                flag=false;
            }
        }
    }
    count+=a[n-1];
    if(flag){
        cout << count <<endl;
    }else{
        cout << -1 <<endl;
    }
    return 0;
}