You work as a system administrator in a dormitory, which has n rooms one after another along a straight hallway. Rooms are numbered from 1 to n.

You have to connect all n rooms to the Internet.

You can connect each room to the Internet directly, the cost of such connection for the i-th room is i coins. 

Some rooms also have a spot for a router. The cost of placing a router in the i-th room is also i coins. You cannot place a router in a room which does not have a spot for it. When you place a router in the room i, you connect all rooms with the numbers from max(1,~i - k) to min(n,~i + k) inclusive to the Internet, where k is the range of router. The value of k is the same for all routers. 

Calculate the minimum total cost of connecting all n rooms to the Internet. You can assume that the number of rooms which have a spot for a router is not greater than the number of routers you have.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of rooms and the range of each router.

The second line of the input contains one string s of length n, consisting only of zeros and ones. If the i-th character of the string equals to '1' then there is a spot for a router in the i-th room. If the i-th character of the string equals to '0' then you cannot place a router in the i-th room.

Output

Print one integer — the minimum total cost of connecting all n rooms to the Internet.

Examples

Input


5 2
00100


Output


3


Input


6 1
000000


Output


21


Input


4 1
0011


Output


4


Input


12 6
000010000100


Output


15

Note

In the first example it is enough to place the router in the room 3, then all rooms will be connected to the Internet. The total cost of connection is 3.

In the second example you can place routers nowhere, so you need to connect all rooms directly. Thus, the total cost of connection of all rooms is 1 + 2 + 3 + 4 + 5 + 6 = 21.

In the third example you need to connect the room 1 directly and place the router in the room 3. Thus, the total cost of connection of all rooms is 1 + 3 = 4.

In the fourth example you need to place routers in rooms 5 and 10. Then all rooms will be connected to the Internet. The total cost of connection is 5 + 10 = 15.
import java.io.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
    
    // static int n;static ArrayList<Integer> arr[];
    // static int m;
    // static boolean v[];
   // int ans[];
    static final long oo=(long)1e18;
    static int mod=1000000007;
    public static void main(String[] args) throws IOException { 
    //    Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        int n=nextInt();
        int k=nextInt();
        String s=next();
   //     pw.println(oo);
        char arr[]=s.toCharArray();
        int j=n-1;
        int closest[]=new int[n];
        int pos=-1;
        Arrays.fill(closest,-1);
        for(int i=n-1;i>=0;i--){
            while(Math.abs(i-j)<=k&&j>=0){
                if(arr[j]=='1'){
                    pos=j;
                   
                }
                j--;
            }
            closest[i]=pos;
        }
        long dp[]=new long[n+1];
        Arrays.fill(dp,oo);
        dp[n]=0;
        for(int i=n-1;i>=0;i--){
            dp[i]=Math.min(dp[i],dp[i+1]+i+1);
            if(closest[i]!=-1){
                int a=Math.max(closest[i]-k,0);
                dp[a]=Math.min(dp[a],dp[i+1]+closest[i]+1);
            }
        }
        pw.println(dp[0]);
        pw.close();
 
    }
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    static class Pair{
        String x;int y;int z;
        Pair(String x,int y,int z){
            this.x=x;
            this.y=y;
            this.z=z;
         //   this.z=z;
          //  this.i=i;
        }
    }
    static class sorting implements Comparator<Pair>{
        public int compare(Pair a,Pair b){
            //return (a.y)-(b.y);
            if(a.y==b.y){
                return -1*(a.z-b.z);
            }
            return (a.y-b.y);
        }
    }
    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // static long modInverse(long a, long m) { 
    //     long g = gcd(a, m); 
    //     if (g != 1) 
    //         return -1; 
    //     else{ 
    //         // If a and m are relatively prime, then modulo inverse 
    //         // is a^(m-2) mode m 
    //         return (power(a, m - 2, m)); 
    //     } 
    // } 
      
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,int M){
        return fast_pow(n,M-2,M);
    }
    
} 

