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;
using Graph = vector<vector<int>>;
#define rep(i, a) for (int i = 0; i < (a); i++ )
#define FOR(i, m, n) for(int i = m; i < n; i++)
 
typedef pair<int,int> P;
typedef long long ll;
 
const int INF = 1001001001;
const int MOD = 1000000007;
const double PI=acos(-1);

ll ans = 0;

bool dfs(int i, int j, vector<int> &B) {
    if (i==0) {
        if (j==0) {
            return true;
        }
        else {
            return false;
        }
    }
    else {
        bool tmp;
        while(1) {
            if (j==B[i-1]+1) {
                ans++;
                B[i] = B[i-1]+1;
                tmp = true;
                break;
            }
            else {
                bool tmp;
                tmp = dfs(i-1,j-1,B);
 
                if (!tmp) {
                    break;
                }
            }
        }
        return tmp;
    }
}

int main(){
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i,N) {
        cin >> A[i];
    }
    vector<int> B(N,0);

    bool tmp;
    rep(i,N) {
        if (A[N-1-i]==B[N-1-i]) {
            continue;
        }
        else {
            tmp = dfs(N-1-i, A[N-1-i], B);

            if (!tmp) {
                break;
            }
        }
    }


    if (tmp) {
        cout << ans << endl;
    }
    else {
        cout << -1 << endl;
    }
}