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 <bits/stdc++.h>
using namespace std;

using i64=int64_t;

#define rep(i,x,y) for(i64 i=i64(x),i##_max_for_repmacro=i64(y); i<i##_max_for_repmacro; ++i)
#define debug(x) #x << "=" << (x)
 
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif
 
const int inf=1.01e9;
const i64 inf64=4.01e18;
const double eps=1e-9;
 
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
    os << "[";
    for (const auto &v : vec) {
        os << v << ",";
    }
    os << "]";
    return os;
}

void solve(){
    i64 N;
    cin >> N;
    vector<i64> A(N);

    rep(i,0,N) cin >> A[i];

    if(A[0]!=0){
        cout << -1 << endl;
        return;
    }
    
    i64 ans=0;

    i64 h=0;
    for(int i=N-1; i>=0; --i){
        --h;
        if(h>A[i]){
            cout << -1 << endl;
            return;
        }
        if(h<A[i]){
            ans+=A[i];
            h=A[i];
        }
    }

    cout << ans << endl;
}

int main(){
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    solve();
    return 0;
}