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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) debug(x) debug(y);
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end() 
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))

template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	//freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
	int n;
	cin >> n;
	vi a(n);
	rep(i, n) cin >> a[i];
	int f = 1;
	if(a[0] != 0){
		cout << -1 << endl;
		return 0;
	}
	int ans = 0;
	int seg = 0;
	rep(i, n){
		if(a[i]){
			if((a[i] - a[i - 1]) > 1){
				f = 0;
				break;
			}
			if(a[i] == a[i - 1]){
			    if(!seg) seg = 2;
			    else seg++;
			}
			else seg = 0;
			if(a[i] == a[i - 1] && a[i] > 1) ans += seg;
			else ans++;
		}
	}
	if(!f){
		cout << -1 << endl;
		return 0;
	}
	cout << ans << endl;
	return 0;
}
/*
Things to look out for:
	- Integer overflows
	- Array bounds
	- Special cases
Be careful!
*/