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 <bits/stdc++.h>

#define foreach(i, v) for(__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define BETWEEN(a, x, b) (a<=x&&x<b)

using namespace std;
typedef long long ll;

ll a[200005];
set<int> st;

int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

  int n;cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];

    if(i>0&&a[i]>a[i-1]+1){
      cout<<-1<<endl;
      return 0;
    }
  }

  if(a[0]>0){
    cout<<-1<<endl;
    return 0;
  }

  ll ans=0;
  ll d=0;

  for(int i=n-1;i>0;i--){
    if(a[i]==0)continue;

    d--;
    st.insert(a[i]-d);
    if(*st.begin() + d == 0){
      st.erase(st.begin());
    }

    ans += st.size();
  }

  cout<<ans<<endl;

  return 0;
}