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<cstdio>
#include<cstring>
#include <cstdlib>  
#include <math.h>
#include <cmath>
#include<cctype>
#include<string>
#include<set>
#include<iomanip>
#include <map>
#include<algorithm>
#include <functional>
#include<vector>
#include<climits>
#include<stack>
#include<queue>
#include<bitset>
#include <deque>
#include <climits>
#include <typeinfo>
#include <utility> 
using namespace std;
using ll = long long;
template<typename T>using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll inf = 1LL << 60;
#define all(x) (x).begin(),(x).end()
#define puts(x) cout << x << endl
#define rep(i,m,n) for(ll i = m;i < n;++i)
#define pb push_back
#define fore(i,a) for(auto &i:a)
#define rrep(i,m,n) for(ll i = m;i >= n;--i)


void err() {
	puts(-1);
	exit(0);
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int N;
	cin >> N;
	vector<int>A(N);
	for (int i = 0; i < N; i++)cin >> A[i];
	int nxt;
	ll ans = 0;
	for (int i = N - 1; i >= 0; i--) {
		if (i == N - 1) {
			ans += A[i];
			if (A[i] > 0)nxt = A[i]-1;
			else nxt = 0;
		}
		else if(i > 0){
			if (nxt != A[i]) {
				if (A[i] > nxt) {
					ans += A[i];
					nxt = A[i] - 1;
				}
				else err();
			}
			else {
				if (A[i] == 0)continue;
				else nxt--;
			}
		}
		else {
			if (nxt != 0)err();
		}
	}
	puts(ans);

	return 0;
}