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: Runtime Error)

#include <cassert>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <numeric>
#include <algorithm>

using namespace std;
using lint = long long;
constexpr int MOD = 1000000007, INF = 1010101010;
constexpr lint LINF = 1LL << 60;

template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
	for (const auto &e : vec) os << e << (&e == &vec.back() ? "" : " ");
	return os;
}

#ifdef _DEBUG
template <class T>
void dump(const char* str, T &&h) { cerr << str << " = " << h << "\n"; };
template <class Head, class... Tail>
void dump(const char* str, Head &&h, Tail &&... t) {
	while (*str != ',') cerr << *str++; cerr << " = " << h << "\n";
	dump(str + (*(str + 1) == ' ' ? 2 : 1), t...);
}
#define DMP(...) dump(#__VA_ARGS__, __VA_ARGS__)
#else 
#define DMP(...) ((void)0)
#endif

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;

	vector<int> A(N);
	for (int i = 0; i < N; i++) cin >> A[i];

	bool ok = A[0] == 0;
	int ans = 0;
	for (int i = N - 1; i >= 0; i--) {

		if (A[i] < 0) {
			ok = false;
			break;
		}
		if (A[i] != 0) {
			ans += A[i];
			for (int j = min(i - A[i], 0); j <= i; j++) {
				A[j] += i - A[i] - j;
			}
		}

	}

	DMP(A);

	if (ok) cout << ans << "\n";
	else cout << -1 << "\n";
	return 0;
}