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 <iostream> // cin, cout, cerr
#include <algorithm> // minmax, sort, swap
#include <numeric> // iota
#include <cstdio> // printf, scanf
#include <string> // string, stoi, to_string
#include <vector> // vector
#include <queue> // queue, priority_queue
#include <deque> // deque
#include <map> // key-value pairs sorted by keys
#include <set> // set
#include <iomanip> // cout<<setprecision(n)
#include <functional> // function<void(int)>

#ifdef DEBUG
#include "debug.hpp"
#else
#define debug(...)
#endif

#define rep(i,n) for(int i = 0; i < (n); i++)
#define ENDL '\n'
#define print(i) std::cout << (i) << '\n'

#define int long long // at least int64 > 9*10^18
#define all(v) (v).begin(), (v).end()
/* libraries */

signed main() {
	int n;
	std::cin >> n;
	std::vector<int> a(n);
	rep(i,n) std::cin >> a[i];
	if(a[0] != 0) {
		print(-1);
		return 0;
	}
	int sum = 0;
	int atleast = 0;
	for(int i=n-1;i>=1;i--) {
		atleast--;
		if(atleast>a[i]) {
			print(-1);
			return 0;
		}
		if(atleast<a[i]) {
			sum+=a[i];
		}
		atleast = a[i];
	}
	print(sum);

	return 0;
}