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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		String s=getString(sc);
      System.out.println(s.replace("?","D"));
      

	}

	/**
	 * *汎用メソッド<br>
	 * 入力した配列に対する順列を取得
	 *
	 * @param numArray
	 * @return
	 */
	private static List<Integer[]> getPermutation(int[] numArray) {
		List<Integer[]> permutation = new ArrayList<>();
		Integer[] ans = new Integer[numArray.length];
		boolean[] used = new boolean[numArray.length];

		return buildPermutaion(permutation, numArray, ans, used, 0);
	}

	/**
	 * *汎用メソッド(内部)<br>
	 * 入力した配列に対する順列を作成
	 *
	 * @param list
	 * @param numArray
	 * @param ans
	 * @param used
	 * @param depth
	 * @return
	 */
	private static List<Integer[]> buildPermutaion(List<Integer[]> list, int[] numArray, Integer[] ans, boolean[] used,
			int depth) {

		// 全部使い切った時=>できた順列を返す
		if (depth == numArray.length) {
			list.add(ans.clone());
			return list;
		}
		// 使い切っていない場合

		for (int i = 0; i < numArray.length; i++) {
			if (used[i]) {
				continue;
			}
			ans[depth] = numArray[i];
			used[i] = true;
			list = buildPermutaion(list, numArray, ans, used, depth + 1);

			used[i] = false;
		}

		return list;
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をlongで返却
	 *
	 * @return
	 */
	public static long getLong(Scanner sc) {

		return Long.parseLong(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をintで返却
	 *
	 * @return
	 */
	public static int getInt(Scanner sc) {

		return Integer.parseInt(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をStringで返却
	 *
	 * @return
	 */
	public static String getString(Scanner sc) {

		return sc.next();
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をint配列で返却
	 *
	 * @return
	 */
	public static int[] getIntArray(Scanner sc) {
		String line = "";
		// 直前にnext()を実行していると改行コードだけ取ってしまうのでその対策
		while (line.equals("")) {
			line = sc.nextLine();
		}

		String[] strArray = line.split(" ");
		int[] intArray = new int[strArray.length];
		for (int i = 0; i < strArray.length; i++) {
			intArray[i] = Integer.parseInt(strArray[i]);
		}
		return intArray;
	}

}