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.util.Scanner;

public class Main {

	public static String Y = "Yes";
	public static String N = "No";
	public static long MOD = (long) (Math.pow(10, 9) + 7);
	public static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {

		char[] t = nc();

		for (char c : t) {
			if (c == '?') {
				outn('D');
			} else {
				outn(c);
			}
		}

		out("");

	}

	public static int count(String search, String text) {
		int nsearch = search.length();
		int ntext = text.length();
		int count = 0;

		for (int i = 0; i <= ntext - nsearch + 1; i++) {
			for (int j = i + nsearch - 1; j < ntext; j++) {
				if (search.equals(text.substring(i, j + 1))) {
					count++;
				}
			}
		}

		return count;
	}

	/*
	 * 以下メソッド集
	 */
	static int ni() {
		return sc.nextInt();
	}

	static long nl() {
		return sc.nextLong();
	}

	static double nd() {
		return sc.nextDouble();
	}

	static String n() {
		return sc.next();
	}

	static char[] nc() {
		return sc.next().toCharArray();
	}

	static int kaijo(int n) {
		if (n == 0 || n == 1) {
			return 1;
		} else {
			return n * kaijo(n - 1);
		}
	}

	static int fib(int n) {
		return (n == 1 || n == 0) ? n : fib(n - 2) + fib(n - 1);
	}

	static long lcm(long m, long n) {

		return m * n / gcd(m, n);
	}

	static int lcm(int m, int n) {

		return m * n / gcd(m, n);
	}

	static long gcd(long a, long b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static void out(String info) {
		System.out.println(info);
	}

	static void out(int info) {
		System.out.println(info);
	}

	static void out(double info) {
		System.out.println(info);
	}

	static void out(long info) {
		System.out.println(info);
	}

	static void out(char info) {
		System.out.println(info);
	}

	static void out(Object obj) {
		System.out.println(obj.toString());
	}

	static void outn(String info) {
		System.out.print(info);
	}

	static void outn(int info) {
		System.out.print(info);
	}

	static void outn(double info) {
		System.out.print(info);
	}

	static void outn(long info) {
		System.out.print(info);
	}

	static void outn(char info) {
		System.out.print(info);
	}

	static int max(int a, int b) {
		return Math.max(a, b);
	}

	static long max(long a, long b) {
		return Math.max(a, b);
	}

	static int min(int a, int b) {
		return Math.min(a, b);
	}

	static long min(long a, long b) {
		return Math.min(a, b);
	}
}

class XY {
	int h;
	int c;

	XY(int h, int c) {
		this.h = h;
		this.c = c;
	}
}