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<string>
#include<algorithm>
#include<utility>
#include<vector>
#include<map>
#include<set>
#include<functional>
#include<iterator>
#include<math.h>

using namespace std;

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

	int n;
	cin >> n;
	int a[n];

	vector<pair<int,int>> idx_val;
	for(int i=0; i<n; i++){
		cin >> a[i];
		if(a[i] > i){
			cout << -1 << "\n";
			return 0;
		}
		if(i != 0 && a[i] - a[i-1] >= 2) {
			cout << -1 << "\n";
			return 0;
		}
		if(i == 0 || a[i-1] >= a[i]) idx_val.emplace_back(make_pair(i,a[i]));
	}

	long long ans=0;
	int last = n;
	for(auto i=idx_val.rbegin();i!=idx_val.rend();i++){
	//	cout << i->first << " " << i->second << "\n";
		int num = i->second-1;
		for(int j=i->first-1;num > 0;j--){
	//		cout << j <<"が"<<a[j]<<"から";
			a[j] -= num;
	//		cout << a[j]<<"に\n";
			ans++;
			num--;
			
		}
		for(int j = i->first; j<last; j++){
			if(a[j] == 0) continue;
	//		cout << j <<"が"<<a[j]<<"から";
			a[j]=0;
	//		cout << a[j]<<"に\n";
			ans++;

		}
		last = i->first;
	}

	cout << ans << "\n";
	return 0;
}