Recently Vasya found a golden ticket — a sequence which consists of n digits a_1a_2... a_n. Vasya considers a ticket to be lucky if it can be divided into two or more non-intersecting segments with equal sums. For example, ticket 350178 is lucky since it can be divided into three segments 350, 17 and 8: 3+5+0=1+7=8. Note that each digit of sequence should belong to exactly one segment.

Help Vasya! Tell him if the golden ticket he found is lucky or not.

Input

The first line contains one integer n (2 ≤ n ≤ 100) — the number of digits in the ticket.

The second line contains n digits a_1 a_2 ... a_n (0 ≤ a_i ≤ 9) — the golden ticket. Digits are printed without spaces.

Output

If the golden ticket is lucky then print "YES", otherwise print "NO" (both case insensitive).

Examples

Input

5
73452


Output

YES


Input

4
1248


Output

NO

Note

In the first example the ticket can be divided into 7, 34 and 52: 7=3+4=5+2.

In the second example it is impossible to divide ticket into segments with equal sum.
import java.util.*;
import java.io.*; 
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void solve(int TC) throws Exception{
        int n = ni();
        String s = n();
        int[] a = new int[n];
        int nn = 0;
        for(int i = 0; i< n; i++)if(s.charAt(i)!='0')a[nn++] = s.charAt(i)-'0';
        if(nn==0)pn("YES");
        else{
            it:for(int l = 1; l<nn;l++){
                int p = 0, sum = 0;
                while(p<l)sum+=a[p++];
                int cur = 0;
                while(p<nn){
                    cur+=a[p++];
                    if(cur>sum)continue it;
                    if(cur==sum)cur = 0;
                }
                if(cur==0){
                    pn("YES");return;
                }
            }
            pn("NO");
        }
    }
    //SOLUTION ENDS
    long mod = (int)1e9+7, IINF = (long)3e18;
    final int MAX = (int)1e6, INF = (int)1e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = true;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}