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 <bits/stdc++.h>

using namespace std;

#define int long long
#define all(v) (v).begin(), (v).end()
#define resz(v, ...) (v).clear(), (v).resize(__VA_ARGS__)
#define reps(i, m, n) for(int i = (int)(m); i < (int)(n); i++)
#define rep(i, n) reps(i, 0, n)

template<class T1, class T2> void chmin(T1 &a, T2 b){if(a>b)a=b;}
template<class T1, class T2> void chmax(T1 &a, T2 b){if(a<b)a=b;}

using Pi = pair<int, int>;
using Tapris = tuple<int, int, int>;
using vint = vector<int>;

const int inf = 1LL << 55;
const int mod = 1e9 + 7;

signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  int n;
  cin >> n;
  vint a(n);
  bool flag = false;
  rep(i, n) {
    cin >> a[i];
    flag |= (i < a[i] || (i && a[i]-a[i-1] > 1));
  }
  if(flag) {
    cout << -1 << endl;
    return 0;
  }
  int ans = n - count(all(a), 0);
  reps(i, 1, n) {
    if(a[i] > 0 && a[i-1] == a[i]) ans += a[i]-1;
  }
  cout << ans << endl;

  return 0;
}