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;

typedef long long int ll;

const ll MAXN = 55556;

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  int n;
  cin >> n;
  vector < int > a(n + 1);
  a[0] = 0;
  for(int i = 1; i <= n; i++) cin >> a[i];
  for(int i = 1; i <= n; i++){
    if(a[i] > i - 1){
      cout << -1 << endl;
      return 0;
    }
  }
  int ans = 0;
  for(int i = n; i >= 1; i--){
    if(a[i] == a[i - 1] + 1) ans++;
    else if(a[i - 1] == a[i]) ans += a[i];
    else if(a[i - 1] < a[i] - 1){
      cout << -1 << endl;
      return 0;
    }else if(a[i - 1] > a[i] && a[i] != 0) ans++;
  }

  cout << ans << '\n';
}