Problem: A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.

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You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.

Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.

Note that the final necklace should remain as one circular part of the same length as the initial necklace.

Input

The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.

Output

Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".

You can print each letter in any case (upper or lower).

Examples

Input

<span class="tex-font-style-tt">-o-o--</span>

Output

YES

Input

<span class="tex-font-style-tt">-o---</span>


Output

YES

Input

<span class="tex-font-style-tt">-o---o-</span>


Output

NO

Input

ooo


Output

YES

Solution (in JAVA):

Solution:
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 p=0,l=0;
		for(char c:s.next().toCharArray())
			if(c=='o')
				++p;
			else
				++l;
		System.out.println(p==0||l%p==0?"YES":"NO");
	}
	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);
	}

}