Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

PD?D??P

Sample Output 1

PDPDPDP

This string contains three occurrences of D and three occurrences of PD as contiguous substrings, so its doctoral and postdoctoral quotient is 6, which is the maximum doctoral and postdoctoral quotient of a string obtained by replacing each ? in T with P or D.

Sample Input 2

P?P?

Sample Output 2

PDPD

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

class B2 {
    public static void main(String... args) {
        final Scanner sc = new Scanner(System.in);
        final char[] T = sc.next().toCharArray();
        final int[] dp = new int[T.length + 1];
        final char[] ans = Arrays.copyOf(T, T.length);
        for (int i = 1; i <= T.length; i++)
            if (T[i - 1] == 'P') {
                dp[i] = dp[i - 1];
            } else if (i >= 2 && T[i - 2] == 'P') {
                dp[i] = dp[i - 2] + 2;
                ans[i - 1] = 'D';
            } else if (i >= 2 && T[i - 2] == '?') {
                if (dp[i - 1] + 1 < dp[i - 2] + 2) {
                    dp[i] = dp[i - 2] + 2;
                    ans[i - 1] = 'P';
                    ans[i - 1] = 'D';
                } else {
                    dp[i] = dp[i - 1] + 1;
                    ans[i - 1] = 'D';
                }
            } else {
                dp[i] = dp[i - 1] + 1;
                ans[i - 1] = 'D';
            }
        System.err.println(dp[T.length]);
        System.out.println(ans);
    }
}
public class Main {
    public static void main(String... args) {
        B2.main();
    }
}