Problem: Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records. 

The records are expressed as a string s of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer p is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer p is considered a period of string s, if for all 1 ≤ i ≤ \lvert s \rvert - p, the i-th and (i + p)-th characters of s are the same. Here \lvert s \rvert is the length of s.

Input

The first line contains two space-separated integers n and p (1 ≤ p ≤ n ≤ 2000) — the length of the given string and the supposed period, respectively.

The second line contains a string s of n characters — Mino's records. s only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that p is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

Input

10 7
1.0.1.0.1.


Output

1000100010


Input

10 6
1.0.1.1000


Output

1001101000


Input

10 9
1........1


Output

No

Note

In the first example, 7 is not a period of the resulting string because the 1-st and 8-th characters of it are different.

In the second example, 6 is not a period of the resulting string because the 4-th and 10-th characters of it are different.

In the third example, 9 is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        FastReader in = new FastReader();
        int n = in.nextInt();
        int p = in.nextInt();
        boolean ans =true;
        char[] str = in.next().toCharArray();
        
        for(int i=0 ;i<n ;i++){
            if(i>=n-p && str[i]=='.') {str[i]='0';continue;}
            if(i>=n-p) continue;
            if(str[i]=='.'){
                if(str[i+p]=='.'){
                    str[i]='0';
                    str[i+p]='1';
                }else str[i] = (str[i+p]=='0')? '1' : '0';
            }
            if(str[i+p]=='.'){
                if(str[i]=='.'){
                    str[i+p]='0';
                    str[i]='1';
                }else str[i+p] = (str[i]=='0')? '1' : '0';
            }
            if(str[i]!=str[i+p]) ans = false;
        }
        
        String s = new String(str);
        if(ans) System.out.println("NO");
        else System.out.println(str);

    // main
    }
    
    
    static class TrieNode{
        Map<Character,TrieNode> children;
        boolean leaf;
        
        TrieNode(){
            leaf = false;
            children = new HashMap<>();
        }
    }
    static TrieNode root = new TrieNode();
    
    static void insert(String key){
        char idx;
        TrieNode mark = root;
        
        for(int lvl=0 ;lvl<key.length() ; lvl++){                
            idx = key.charAt(lvl);
            

            if(mark.children.get(idx)==null) mark.children.put(idx,new TrieNode());
            
            mark = mark.children.get(idx);
        }
        mark.leaf = true;
    }
    
    static boolean search(String key){
        char idx;
        TrieNode mark = root;
        
        for(int lvl=0 ;lvl<key.length() ; lvl++){
            idx = key.charAt(lvl);
            
            if(mark.children.get(idx)==null)
                return false;
            mark = mark.children.get(idx);
        }
        if(mark!=null && mark.leaf) return true;
        else return false;
    }
    
    
    
    //FR
    static class FastReader  {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}