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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Iterator;
import java.util.PrimitiveIterator;
import java.util.Scanner;
import java.util.function.Supplier;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main{
	static Scanner s=new Scanner(System.in);

	void solve(){
		int n=gInt();
		int[]a=INTS(n).toArray();
		if(a[0]>0) {
			System.out.println(-1);
			return;
		}
		
		long r=0;
		for(int i:rep(1,n-1)) {
			if(a[i]==a[i-1]+1)
				++r;
			else
				r+=a[i];
		}
		System.out.println(r);
	}

	public static void main(String[] A){
		new Main().solve();
	}

	static int gInt(){
		return Integer.parseInt(s.next());
	}
	static long gLong(){
		return Long.parseLong(s.next());
	}
	static double gDouble(){
		return Double.parseDouble(s.next());
	}

	SupplyingIterator<Integer> ints(int n){
		return new SupplyingIterator<>(n,Main::gInt);
	}
	SupplyingIterator<Long> longs(int n){
		return new SupplyingIterator<>(n,Main::gLong);
	}
	SupplyingIterator<Double> doubles(int n){
		return new SupplyingIterator<>(n,Main::gDouble);
	}
	SupplyingIterator<String> strs(int n){
		return new SupplyingIterator<>(n,s::next);
	}

	Range rep(int i){
		return Range.rep(i);
	}
	Range rep(int f,int t,int d){
		return Range.rep(f,t,d);
	}
	Range rep(int f,int t){
		return rep(f,t,1);
	}
	Range rrep(int f,int t){
		return rep(t,f,-1);
	}

	IntStream REP(int v){
		return IntStream.range(0,v);
	}
	IntStream REP(int l,int r){
		return IntStream.rangeClosed(l,r);
	}

	IntStream INTS(int n){
		return IntStream.generate(Main::gInt).limit(n);
	}
	Stream<String> STRS(int n){
		return Stream.generate(s::next).limit(n);
	}

}
class SupplyingIterator<T> implements Iterable<T>,Iterator<T>{
	int			t;
	Supplier<T>	supplier;

	SupplyingIterator(int t,Supplier<T> supplier){
		this.t=t;
		this.supplier=supplier;
	}

	@Override
	public Iterator<T> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return t>0;
	}

	@Override
	public T next(){
		--t;
		return supplier.get();
	}

}
class Range implements Iterable<Integer>,PrimitiveIterator.OfInt{
	int to,cur,d;

	Range(int from,int to,int d){
		this.cur=from-d;
		this.to=to;
		this.d=d;
	}

	Range(int n){
		this(0,n-1,1);
	}

	@Override
	public Iterator<Integer> iterator(){
		return this;
	}

	@Override
	public boolean hasNext(){
		return cur+d==to||(cur!=to&&(cur<to==cur+d<to));
	}

	@Override
	public int nextInt(){
		return cur+=d;
	}

	static Range rep(int i){
		return new Range(i);
	}
	static Range rep(int f,int t,int d){
		return new Range(f,t,d);
	}
	static Range rep(int f,int t){
		return rep(f,t,1);
	}
	static Range rrep(int f,int t){
		return rep(f,t,-1);
	}

}