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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <string>
#include <vector>
#include <limits>
#include <numeric>
#include <queue>
#include <cmath>
#include <set>
#include <map>

using namespace std;

#define INF (1ll<<60)
#define INFint (1<<30)
#define MOD 1000000007
#define BOUND 27182818284

typedef long long ll;
typedef long long int lli;
typedef pair<ll, ll> P;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i, a, b) for(int i=int(a);i<int(b);++i)

template<class T>
bool umax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool umin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

// gcd
template<typename T>
T gcd(T a, T b) {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int findGCD(int arr[], int n) {
    int result = arr[0];
    for (int i = 1; i < n; i++)
        result = gcd(arr[i], result);
    return result;
}

template<typename T>
T lcm(T m, T n) {
    // 引数に0がある場合は0を返す
    if ((0 == m) || (0 == n))
        return 0;
    return ((m / gcd(m, n)) * n); // lcm = m * n / gcd(m,n)
}

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    fill((T *) array, (T *) (array + N), val);
}


int dx[5] = {1, 0, -1, 0};
int dy[5] = {0, 1, 0, -1};

// v.front() = -BOUND;
// v.back() = BOUND;

//struct edge{
//    int cost, to;
//
//    edge(int in_cost, int in_to){
//        cost=in_cost;
//        to=in_to;
//    }
//    bool operator<(const edge &a) const
//    {
//        return cost > a.cost;
//    }
//};

int main() {
    int N; cin >> N;
    vector<int>A(N,0);

    rep(i,N){
        cin >> A[i];
    }

    if(A[0]!=0){
        cout << -1 << endl;
        return 0;
    }

    lli ans=0LL;
    for(int i=0; i<N-1; i++){
        if(A[i]+1==A[i+1]){
            ans++;
        }else if(A[i]>=A[i+1] && A[i+1]>1){
            ans+=2LL;
        }else if(A[i+1]==1) {
            ans++;
        }else if(A[i]+1<A[i+1]){
            cout << -1 << endl;
        }
    }

    cout << ans << endl;
    return 0;
}