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.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

import javax.net.ssl.SSLContext;

public class Main
{
   static long mod=(long)(1e+9) + 7;
  
   static int[] sieve;
   static ArrayList<Integer> primes;

   public static void main(String[] args) throws java.lang.Exception
   {	  	
	     fast s = new fast();
	     PrintWriter out=new PrintWriter(System.out);
	     Scanner sc=new Scanner(System.in);
	     StringBuilder final_ans = new StringBuilder();      
	     
	     int n=s.nextInt();
	     int p=s.nextInt();
	     
	     String strt=s.nextLine();
	     StringBuilder str=new StringBuilder(strt);
	     StringBuilder str1=new StringBuilder(strt);
	     
	     int flag=1;
	     
	     for(int i=0;i<str1.length()-p;i++)
	     {
	    	 if(str1.charAt(i)!='.' && str1.charAt(i+p)=='.')
	    	 {
	    		 if(str1.charAt(i)=='0') {str1.deleteCharAt(i+p);str1.insert(i+p, '1');}
	    		 else {str1.deleteCharAt(i+p);str1.insert(i+p, '0');}
	    		 flag=0;
	    	 }
	    	 else if(str1.charAt(i)!='.' && str1.charAt(i+p)!='.')
	    	 {
	    		 char c=str1.charAt(i);
	    		 char d=str1.charAt(i+p);
	    		 if(c!=d) flag=0;
	    	 }
	    	 else if(str1.charAt(i)=='.' && str1.charAt(i+p)=='.')
	    	 {
	    		 str1.deleteCharAt(i);	str1.insert(i, '0');
	    		 str1.deleteCharAt(i+p); str1.insert(i+p, '1');
	    		 flag=0;
	    	 }
	    	 else if(str1.charAt(i)=='.' && str1.charAt(i+p)!='.')
	    	 {
	    		 char c=str1.charAt(i+p);
	    		 if(c=='1') {str1.deleteCharAt(i);str1.insert(i, '0');}
	    		 else {str1.deleteCharAt(i);str1.insert(i, '1');}
	    		 flag=0;
	    	 }
	     }
	     
	     for(int i=0;i<str1.length() && flag==0;i++)
	     {
	    	 if(str1.charAt(i)=='.') {str1.deleteCharAt(i); str1.insert(i,'0');}
	     }
	     if(flag==1) System.out.println("No");
	     else System.out.println(str1);
   }
   
   

   static class fast {
		private InputStream i;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		
		public static void sieve(int size)
		{
			sieve=new int[size];
			primes=new ArrayList<Integer>();
			sieve[1]=1;
			for(int i=2;i<=Math.sqrt(size);i++)
			{
				if(sieve[i]==0)
				 {
					for(int j=i*i;j<size;j+=i) sieve[j]=1;
				 }
			}
			
			for(int i=2;i<size;i++)
			{
				if(sieve[i]==0) primes.add(i);
			}
		}
		public static long pow(long n, int b, long MOD)
		   {
		       long x=1;long y=n; 
		       while(b > 0)
		       {
		           if(b%2 == 1)
		           {
		               x=x*y;
		               if(x>MOD) x=x%(MOD);
		           }
		           y = y*y;
		           if(y>MOD) y=y%(MOD); 
		           b >>= 1;
		       }
		       return x;
		   }

		
		 public static int lower(Integer[] a,int start,int end,int key)
		 {
			   int mid=(start+end)>>1;
			   if(start==end && a[mid]<key) {return -1;}
			   if(start>end) return -1;

			   if(a[mid]>=key && (((mid-1)>=0 && a[mid-1]<key) || (mid-1)==0)) return mid;
			   else if(a[mid]== key && (mid-1)>=0 && a[mid-1]==key) return lower(a,start,mid-1,key);   
			   else if(key>a[mid]) return lower(a,mid+1,end,key);
			   else return lower(a,start,mid-1,key);
		}
		 
		 public static int upper(Integer a[],int start,int end,int key)
		 {
			   int mid=(start+end)>>1;
			   if(start==end && a[mid]>key) {return -1;}
			   if(start>end) return -1;

			   if(a[mid]<=key && (((mid+1)<a.length && a[mid+1]>key) || (mid+1)==a.length)) return mid;
			   else if(a[mid]== key && (mid+1)<a.length && a[mid+1]==key) return upper(a,mid+1,end,key);   
			   else if(key>=a[mid]) return upper(a,mid+1,end,key);
			   else return upper(a,start,mid-1,key);
		} 
		   
		public int gcd(int a,int b)
		{
			if(a==0) return b;
			return gcd(b%a,a);
		}
		public fast() {
			this(System.in);
		}
		public fast(InputStream is) {
			i = is;
		}
		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = i.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}
		public String nextString() {
     		int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		public long nextLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public int nextInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}	
}