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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Pradyumn
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, FastReader in, OutputWriter out) {
            int n = in.nextInt();
            int p = in.nextInt();
            char[] s = in.nextCharacterArray(n);
            boolean rem = false;
            for (int i = 0; i + p < n; ++i) {
                if (s[i] != '.' && s[i + p] != '.') {
                    rem |= s[i] != s[i + p];
                } else if (s[i] == '.' || s[i + p] == '.') {
                    rem = true;
                    if (s[i] == '.' && s[i + p] != '.') {
                        int cur = s[i + p] - '0';
                        cur ^= 1;
                        s[i] = (char) (cur + '0');
                    } else if (s[i + p] == '.' && s[i] != '.') {
                        int cur = s[i] - '0';
                        cur ^= 1;
                        s[i + p] = (char) (cur + '0');
                    } else {
                        s[i] = '0';
                        s[i + p] = '1';
                    }
                }
            }
            if (rem) {
                for (int i = 0; i < p; ++i) if (s[i] == '.') s[i] = '1';
                out.printLine(new String(s));
                return;
            } else if (p == n) {
                out.printLine("No");
            } else {
                for (int i = 0; i < n; ++i) if (s[i] == '.') s[i] = '1';
                boolean isP = n > p;
                for (int i = 0; i + p < n; ++i) isP &= s[i] == s[i + p];
                if (isP) out.printLine("No");
                else out.printLine(new String(s));
            }
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void print(Object... objects) {
            for (int i = 0; i < objects.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(objects[i]);
            }
        }

        public void printLine(Object... objects) {
            print(objects);
            writer.println();
        }

        public void close() {
            writer.close();
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        private int pread() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = pread();
            }
            int res = 0;
            do {
                if (c == ',') {
                    c = pread();
                }
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = pread();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public char nextCharacter() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            return (char) c;
        }

        public char[] nextCharacterArray(int n) {
            char[] chars = new char[n];
            for (int i = 0; i < n; i++) {
                chars[i] = nextCharacter();
            }
            return chars;
        }

    }
}