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.util.*;

import javax.swing.plaf.synth.SynthSpinnerUI;

import java.io.*;

public class Main {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		//DisjointSet set = new DisjointSet(26);
		
		InputReader sc1 = new InputReader(System.in);
		PrintWriter sc2 = new PrintWriter(System.out);
		int n = sc1.nextInt();
		int p = sc1.nextInt();
		String s = sc1.readString();
		char c[] = s.toCharArray();
		int count = 0;
		boolean flag = false;
		for(int i=0;i<n;i++) {
			if(i+p==n) {
				break;
			}
			if(c[i]=='1') {
				if(c[i+p]!='1') {
					c[i+p] = '0';
					flag = true;
				}
			}
			else if(c[i]=='0') {
				if(c[i+p]!='0') {
					c[i+p] = '1';
					flag = true;
				}
			}
			else {
				if(c[i+p]=='1') {
					c[i] = '0';
				}
				else if(c[i+p]=='0') {
					c[i] = '1';
				}
				else {
					c[i] = '0';
					c[i+p] = '1';
				}
				flag = true;
			}
		}
		if(flag==false) {
			System.out.println("NO");
		}else {
			for(int i=0;i<n;i++) {
				if(c[i]=='.') {
					c[i] = '0';
				}
				System.out.print(c[i]);
			}
		}
		sc2.close();
	}

	/*
	 * 	DIVISORS
	 * */
	public static void divsors(int n) {
		for(int i=1;i<=Math.sqrt(n)+1;i++) {
			if(n%i==0) {
				if(n/i==i) {
					System.out.print(i+" ");
				}
				else {
					System.out.print	(i+" "+n/i);
				}
			}
		}
	}
	
	/*
	 * 	GCD 
	 * */
	
	public static int GCD(int x, int y) {
		if(x==0) {
			return y;
		}
		return GCD(y%x, x);
	}
	
	/*
	 * 	GCD EXTENDED
	 * */
	
	public static int gcdExtended(int a, int b, int x, int y) {
		if(a==0) {
			x = 0;
			y = 1;
			return b;
		}
		
		int x1 = 1, y1 = 1;
		int gcd = gcdExtended(b%a, a, x1, y1);
		x = y1 - (b/a)*x1;
		y = x1;
		
		return gcd;
	}
	
	
	static class DisjointSet{
		static StringBuilder str = new StringBuilder();
		static int count = 0;
		int rank[];
		int parent[];
		int n;
		DisjointSet(int n){
			this.n = n;
			rank = new int[n];
			Arrays.fill(rank, 0);
			parent = new int[n];
			makeSet();
			
		}
		void makeSet() {
			for(int i=0;i<n;i++) {
				parent[i] = i;
			}
		}
		
		int find(int x) {
			if(parent[x]!=x) {
				parent[x] = find(parent[x]);
			}
			return parent[x];
		}
		
		void union(int x, int y) {
			int xRoot = find(x);
			int yRoot = find(y);
			
			if(xRoot==yRoot) {
				return;
			}
			
			if(rank[xRoot]>rank[yRoot]) {
				parent[yRoot] = xRoot;
				Append(x,y);
				count++;
			}
			else if(rank[xRoot]<rank[yRoot]) {
				parent[xRoot] = yRoot;
				Append(x,y);
				count++;
			}
			else {
				parent[yRoot] = xRoot;
				rank[xRoot] = rank[xRoot] + 1;
				count++;
				Append(x,y);
			}
		}
		
		void Append(int x, int y) {
			str.append((char)(x+97)+" "+(char)(y+97)+"\n");
		}
	}
	static class InputReader {
		 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
		private SpaceCharFilter filter;
 
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
 
		public int snext() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public int nextInt() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nextLong() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nextIntArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
 
		public String readString() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = snext();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}	
}