You are given a huge decimal number consisting of n digits. It is guaranteed that this number has no leading zeros. Each digit of this number is either 0 or 1.

You may perform several (possibly zero) operations with this number. During each operation you are allowed to change any digit of your number; you may change 0 to 1 or 1 to 0. It is possible that after some operation you can obtain a number with leading zeroes, but it does not matter for this problem.

You are also given two integers 0 ≤ y < x < n. Your task is to calculate the minimum number of operations you should perform to obtain the number that has remainder 10^y modulo 10^x. In other words, the obtained number should have remainder 10^y when divided by 10^x.

Input

The first line of the input contains three integers n, x, y (0 ≤ y < x < n ≤ 2 ⋅ 10^5) — the length of the number and the integers x and y, respectively.

The second line of the input contains one decimal number consisting of n digits, each digit of this number is either 0 or 1. It is guaranteed that the first digit of the number is 1.

Output

Print one integer — the minimum number of operations you should perform to obtain the number having remainder 10^y modulo 10^x. In other words, the obtained number should have remainder 10^y when divided by 10^x.

Examples

Input


11 5 2
11010100101


Output


1


Input


11 5 1
11010100101


Output


3

Note

In the first example the number will be 11010100100 after performing one operation. It has remainder 100 modulo 100000.

In the second example the number will be 11010100010 after performing three operations. It has remainder 10 modulo 100000.
import java.util.*;
import java.io.*;
import java.math.*;
public class loser
{
    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;
        public InputReader(InputStream stream)
        {
            br=new BufferedReader(new InputStreamReader(stream),32768);
            token=null;
        }

        public String next()
        {
            while(token==null || !token.hasMoreTokens())
            {
                try
                {
                    token=new StringTokenizer(br.readLine());
                }
                catch(IOException e)
                {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int nextInt()
        {
            return Integer.parseInt(next());
        }

        public long nextLong()
        {
            return Long.parseLong(next());
        }
    }
    static class card{
        long g;
        int ess;
        public card(long ch,int i)
        {
            this.g=ch;
            this.ess=i;
        }
    }
    static class sort implements Comparator<card>
    {
        public int compare(card o1,card o2)
        {
            if(o1.ess!=o2.ess)
                return (o1.ess-o2.ess);
            else
                return (int)(o1.g-o2.g);
        }
    }
    /*static void shuffle(long a[])
    {
    List<Long> l=new ArrayList<>();
    for(int i=0;i<a.length;i++)
    l.add(a[i]);
    Collections.shuffle(l);
    for(int i=0;i<a.length;i++)
    a[i]=l.get(i);
    }

    static long gcd(long a,long b)
    {
    if(b==0)
    return a;
    else
    return gcd(b,a%b);
    }
    /*static int ans1=Integer.MAX_VALUE,ans2=Integer.MAX_VALUE,ans3=Integer.MAX_VALUE,ans4=Integer.MAX_VALUE;
    static boolean v[]=new boolean[101];
    static void dfs(Integer so,Set<Integer> s[]){
    if(!v[so.intValue()])
    {
    v[so]=true;
    for(Integer h:s[so.intValue()])
    {
    if(!v[h.intValue()])
    dfs(h,s);
    }
    }
    }
    static class Print{
    public PrintWriter out;
    Print(OutputStream o)
    {
    out=new PrintWriter(o);
    }
    }
    static int CeilIndex(int A[], int l, int r, int key) 
    { 
    while (r - l > 1) { 
    int m = l + (r - l) / 2; 
    if (A[m] >= key) 
    r = m; 
    else
    l = m; 
    } 

    return r; 
    } 

    static int LongestIncreasingSubsequenceLength(int A[], int size) 
    { 
    // Add boundary case, when array size is one 
    int[] tailTable = new int[size]; 
    int len; // always points empty slot 
    tailTable[0] = A[0]; 
    len = 1; 
    for (int i = 1; i < size; i++) { 
    if (A[i] < tailTable[0]) 
    // new smallest value 
    tailTable[0] = A[i]; 

    else if (A[i] > tailTable[len - 1]) 
    // A[i] wants to extend largest subsequence 
    tailTable[len++] = A[i]; 

    else
    // A[i] wants to be current end candidate of an existing 
    // subsequence. It will replace ceil value in tailTable 
    tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i]; 
    } 
    return len; 
    }*/
    /*static int binary(int n)
    {
    int s=1;
    while(n>0)
    {
    s=s<<1;
    n--;
    }
    return s-1;
    }
    static StringBuilder bin(int i,int n)
    {
    StringBuilder s=new StringBuilder();
    while(i>0)
    {
    s.append(i%2);
    i=i/2;
    }
    while(s.length()!=n)
    {
    s.append(0);
    }
    return s.reverse();
    }*/
    /*static boolean v[]=new boolean[1000006];
    static void seive()
    {
    for(int i=2;i<1000006;i++)
    {
    if(v[i]==false)
    {
    for(int j=2*i;j<1000006;j+=i)
    {
    v[j]=true;
    }
    }
    }
    }*/
    public  static void main(String[] args) 
    {
        InputReader sc=new InputReader(System.in);
        int n=sc.nextInt();
        int x=sc.nextInt();
        int y=sc.nextInt();
        char c[]=sc.next().toCharArray();
        int ans=0;
        for(int i=n-x;i<n;i++)
        {
            if(i!=(n-(y+1)) && c[i]!='0')
            ans++;
            if(i==(n-(y+1)) && c[i]!='1')
            ans++;
        }
        System.out.println(ans);
    }
}