Problem: For a string S consisting of the uppercase English letters `P` and `D`, let the doctoral and postdoctoral quotient of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3.

We have a string T consisting of `P`, `D`, and `?`.

Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.

Constraints

* 1 \leq |T| \leq 2 \times 10^5
* T consists of `P`, `D`, and `?`.

Input

Input is given from Standard Input in the following format:


T


Output

Print one string with the maximum possible doctoral and postdoctoral quotient among the strings that can be obtained by replacing each `?` in T with `P` or `D`. If there are multiple such strings, you may print any of them.

Examples

Input

PD?D??P


Output

PDPDPDP


Input

P?P?


Output

PDPD

Solution (in JAVA):

Solution:

import java.text.DecimalFormat;
import java.util.stream.LongStream;
import java.util.stream.IntStream;
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        AtCoder problem = new AtCoder(sc);
        problem.solve(out);

        out.flush();
    }

}

class AtCoder {

    final String T;

    AtCoder(FastScanner sc) {
        T = sc.next();
    }

    void solve(PrintWriter out) {
        String sd = T.replace('?', 'D');
        out.println(sd);
    }

}

// https://qiita.com/p_shiki37/items/a0f6aac33bf60f5f65e4
class FastScanner {

    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte()) {
            return buffer[ptr++];
        } else {
            return -1;
        }
    }

    private static boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
            ptr++;
        }
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) {
            throw new NumberFormatException();
        }
        return (int) nl;
    }

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