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 <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define rep(i,s,e) for(int (i) = (s);(i) <= (e);(i)++)
#define all(x) x.begin(),x.end()

int N;
vector<i64> A;
int main(){
  cin >> N;
  A.resize(N);
  rep(i,0,N - 1){
    cin >> A[i];
  }

  if(A[0] != 0){
    cout << -1 << endl;
    return 0;
  }
  if(A[N - 1] > N - 1){
    cout << -1 << endl;
  }
  i64 ans = A[N - 1];
  for(int i = N - 1;i >= 1;i--){
    if(A[i] < A[i - 1]){
      if(A[i - 1] > i - 1){
        cout << -1 << endl;
      }
      ans += A[i - 1];
    }
    else if(A[i] - A[i - 1] == 0){
      if(A[i - 1] > i - 1){
        cout << -1 << endl;
      }
      ans += A[i - 1];
    }
    else if(A[i] - A[i - 1] == 1){
    }
    else{
      cout << -1 << endl;
    }
  }

  cout << ans << endl;
}