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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class CodeA {
	public static void main(String args[]) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		StringBuilder sb = new StringBuilder();
		// ----------My Code----------

		int n = in.nextInt();
		int p = in.nextInt();

		char str[] = in.next().toCharArray();

		boolean ans = true;
		for (int i = 0; i < n; i++) {
			if (str[i] == '.') {
				if (i + p < n) {
					if (str[i + p] == '.') {
						str[i] = '0';
						str[i + p] = '1';
					} else {
						str[i] = (char) (1 - ((int) str[i + p] - 48) + 48);
					}
				} else if (i - p >= 0) {
					str[i] = (char) (1 - ((int) str[i - p] - 48) + 48);
				} else {
					str[i] = '0';
				}
			}
		}

		for (int i = 0; i < n; i++) {
			if (i + p < n) {
				if (str[i] != str[i + p]) {
					ans = false;
					break;
				}
			}
		}

		if (ans == true) {
			System.out.println("No");
		} else {
			System.out.println(str);
		}

		// ---------------The End------------------

		out.close();
	}

	// ---------------Extra Methods------------------

	public static long pow(long x, long n, long mod) {
		long res = 1;
		x %= mod;
		while (n > 0) {
			if (n % 2 == 1) {
				res = (res * x) % mod;
			}
			x = (x * x) % mod;
			n /= 2;
		}
		return res;
	}

	public static long gcd(long x, long y) {
		if (x % y == 0)
			return y;
		else
			return gcd(y, x % y);
	}

	// ---------------Extra Methods------------------

	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine() {
			String fullLine = null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine = reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}

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

		public char nextChar() {
			return next().charAt(0);
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

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

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n, int f) {
			if (f == 0) {
				int[] arr = new int[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextInt();
				}
				return arr;
			} else {
				int[] arr = new int[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextInt();
				}
				return arr;
			}
		}

		public long[] nextLongArray(int n, int f) {
			if (f == 0) {
				long[] arr = new long[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextLong();
				}
				return arr;
			} else {
				long[] arr = new long[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextLong();
				}
				return arr;
			}

		}

		public double[] nextDoubleArray(int n, int f) {
			if (f == 0) {
				double[] arr = new double[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			} else {
				double[] arr = new double[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			}
		}

	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			if (this.a != o.a)
				return Integer.compare(this.a, o.a);
			else
				return Integer.compare(this.b, o.b);
		}

		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.a == a && p.b == b;
			}
			return false;
		}

		public int hashCode() {
			return new Integer(a).hashCode() * 31 + new Integer(b).hashCode();
		}

	}

}