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.io.PrintWriter;

import java.util.*;
import java.util.Arrays ; 
import java .lang.String.* ;
import java .lang.StringBuilder ;


public class Test{
    static int pos = 0 ; 
    static int  arr[] ; 
  static LinkedList l1 = new  LinkedList() ; 
static void find(int p ,char[]x,int put[],String s){
    int c= 0 ; 
    for (int i = 0; i < s.length(); i++) {
        if(x[p]==s.charAt(i)){
        c++ ; }
    }
    put[p] = c ;  
}
static int mode(int m ,int[]x ){
    int temp = 0 ; 
    for (int i = x.length-1; i >=0; i--) {
        if(x[i]<=m){
            temp= x[i] ; 
           /// break ; 
             return m-temp ; 
             
        }
    }
     return m-temp ; 
}
static int mode2(int m ,int[]x ){
    int temp = 0 ; 
    
    for (int i = x.length-1; i >=0; i--) {
        if(x[i]<=m){
            temp= x[i] ; 
           /// break ; 
             return x[i] ; 
             
        }
    }
     return 0 ; 
}
static int find(int x[],int temp){
    int j = 0 ; 
    for (int i = x.length-1; i >=0; i--) {
        if(x[i]==temp) return j+1 ; 
        j++ ; 
    }
    return -1 ; 
}
static String ch(long[]x,long b){
    for (int i = 0; i < x.length; i++) {
        if(x[i]==b)return "YES" ;
    }
    return "NO" ; 
}
static int pis(String s,char k ){
  
    for (int i = 0; i < s.length(); i++) {
        if(k==s.charAt(i) ) {
            return i ;
        } 
    }
   
    return -1 ; 
    
}
static  int fin(String s){
   int count = 0 ; 
    for (int i = 0; i < s.length(); i++) {
       if(s.charAt(i)>=65&&s.charAt(i)<=90) count++ ; 
    }
   return count ; 
}
static boolean config(int[]x ){
    for (int i = 0; i < x.length; i++) {
        if(x[i]==0) return false ;
    }
    return true  ; 
}
static long count(long i ){
    if(i%2==0){
        return i ; 
    }else return i+count(i-1) ;  
}
static boolean ce(int [][]x){
    for (int i = 0; i < x.length; i++) {
        if(x[i][0]!=x[i][1]) return true ; 
    }
    
    return false ;
}
static boolean c(int[][]x,int[]y){
    int j=y.length-1 ; 
    for (int i = 0; i < x.length; i++) {
        if(x[i][1]!=y[j]) return true;
        j-- ;
    }
    return false  ; 
}
static long gcd(long a , long b){
    if(b==0) return a ; 
    else return gcd(b,a%b) ; 
    
}
static boolean prime(int n,int y){
    int count =0 ; 
    for (int i = 2; i <= Math.max(n, y); i++) {
        if(n%i==0&&y%i==0)count++ ; 
    }
    if(count>0) return false ;
    
    return true ;
}
static int getbig(int []x,int r){
    for (int i = 0; i < x.length; i++) {
        if(x[i]==r) return i ;
    }
    return -1 ; 
}
static int getsmaal(int []x,int r){
    for (int i = x.length-1; i >= 0; i--) {
        if(x[i]==r) return i ;
    }
    return -1 ; 
}
static void find(LinkedList l1 ,int y[]){
    int v[] =new int[3] ; 
  
    for (int i = 1; i <= 3; i++) {

        for (int j = 0; j < y.length; j++) {
          
            if(y[j]==i&&!l1.contains(j)){
             v[i-1]=j+1 ; 
             l1.add(j) ;
            // count++ ;
             break ; 
            }
        }
        
    }
    System.out.println(v[0]+" "+v[1]+" "+v[2]+" ");
}
    public static void main(String[] args)  {

        Scanner in = new Scanner(System.in) ;
        PrintWriter pw = new PrintWriter(System.out);  
      
        String s= in.next() ;
        int o = 0 ;
        int v = 0 ;
        for (int i = 0; i < s.length(); i++) {
            if(s.charAt(i)=='o'){
                o++ ;
            } else v ++ ; 
        }
        if(o==0){
            System.out.println("YES");
            return ;
        }
        if(v%o==0)System.out.println("YES");else System.out.println("NO");
                pw.close();
    }
}