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 C# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

//
// Nomura2020B
// URL : https://atcoder.jp/contests/nomura2020/tasks/nomura2020_b
// 
using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;

namespace AtCoder
{
    class Nomura2020B
    {
        public static readonly long MOD_CONST = 1000000007;//(long)Math.Pow(10, 9) + 7;


        static void Main(string[] args)
        {
            //出力高速化
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);
            Solve();
            Console.Out.Flush();
        }
        public enum PDCase
        {
            CaseP = 0,
            CaseD = 1,
            CaseX = 2, 
        };
        public static void Solve()
        {
            string T = Console.ReadLine();
            var TC = new PDCase[T.Length];

            for(int i = 0; i < T.Length; ++i)
            {
                switch (T[i])
                {
                    case 'P': TC[i] = PDCase.CaseP; break;
                    case 'D': TC[i] = PDCase.CaseD; break;
                    case '?': TC[i] = PDCase.CaseX; break;
                }
            }
            string ansStr = "";
            for (int i = 0; i < TC.Length - 1; ++i)
            {
                if (TC[i] == PDCase.CaseX)
                {
                    switch (TC[i + 1])
                    {
                        case PDCase.CaseP:
                            TC[i] = PDCase.CaseD;
                            break;
                        case PDCase.CaseD:
                            TC[i] = PDCase.CaseP;
                            break;
                        case PDCase.CaseX:
                            TC[i] = PDCase.CaseP;
                            TC[i + 1] = PDCase.CaseD;
                            break;
                    }
                }
            }
            foreach (var x in TC)
            {
                ansStr += (x == PDCase.CaseP) ? "P" : "D";
            }
            Console.WriteLine(ansStr);
        }
#if false
        public static int calc(PDCase[] ans)
        {
            int x = 0;
            for (int i = 1; i < ans.Length; ++i)
            {
                if (ans[i] == PDCase.CaseD)
                {
                    if (ans[i - 1] == PDCase.CaseP)
                    {
                        x += 2;//PD, D
                    }
                    ++x;//D
                }
            }
            return x;
        }
#endif
        //---------------------------------------------------------------------------------
#region MyLib

        /// <summary>
        /// グラフ系
        /// </summary>
        public class Node
        {
            public Node(long val){
                this.Value = val;
                this.Children = new List<Node>();
            }

            public Node(long val, List<Node> children)
            {
                this.Value = val;
                this.Children = children;
            }

            public long Value { get; set; }
            public List<Node> Children { get; set; }
        }

        /// <summary>
        /// Clamping
        /// </summary>
        /// <typeparam name="T">比較型</typeparam>
        /// <param name="x">clampしたい値</param>
        /// <param name="min">最小値</param>
        /// <param name="max">最大値</param>
        /// <returns>clamp済みの値</returns>
        public static T Clamp<T>(T x, T min, T max) where T : IComparable<T>
        {
            if (x.CompareTo(min) < 0) { return min; }
            else if (x.CompareTo(max) > 0) { return max; }
            else return x;
            //return Math.Max(Math.Min(x, max), min);
        }

        /// <summary>
        /// 指定された底log_baseでの, 対数を計算
        /// </summary>
        /// <param name="x"></param>
        /// <param name="log_base">指定の底</param>
        /// <returns></returns>
        public static double LogBase(double x, double log_base)
        {
            //log_b(N) = log_a(N) / log_a(B)より
            return Math.Log(x) / Math.Log(log_base);
        }

        /// <summary>
        /// 初項a, 公差d, 項数nの数列の和
        /// </summary>
        /// <param name="n">求めたいところ</param>
        /// <returns>初項からxまで足した値</returns>
        static long SumSeries(long a, long d, long n)
        {
            return (2 * a + (n - 1) * d) * n / 2;
        }
        /// <summary>
        /// 累積和の配列を計算する
        /// Sum[ a[0], a[i] )がs[i+1]に入っている
        /// s[0]  : 0                //一切和を取らない場合
        /// s[1]  : ls[0]            //a[0]   + s[0] = a[0]
        /// s[2]  : ls[1] + s[1]     //a[1]   + a[0]
        /// ...
        /// s[N]  : ls[N-1] + s[N-1] //a[N-1] + ... + a[0]
        /// s[N+1]: ls[N] + s[N]     //a[N]   + ... + a[0]
        /// 
        /// 例: [1, 7)の和(a[1] + ... + a[6])
        /// s[1] = a[0] //[0, 1)
        /// s[7] = a[0] + a[1] + ... + a[6] //[0, 7)
        /// よってs[7] - s[1]);
        /// </summary>
        /// <param name="ls">計算したい配列</param>
        /// <returns>累積和配列</returns>
        static List<long> AggregateSumList(List<long> a)
        {
            var s = new List<long>(a.Count + 1);
            if (a.Count == 0)
            {
                return s;
            }
            s.Add(0);//s[0] = 0
            for (int i = 0; i < a.Count; ++i)
            {
                s.Add(a[i] + s[i]);
            }
            return s;
        }


        /// <summary>
        /// 与えられた数が素数かどうかを計算する
        /// </summary>
        /// <param name="num">対象数</param>
        /// <returns>true:素数, false:素数ではない(合成数)</returns>
        public static bool isPrime(int num)
        {
            if (num < 2) return false;
            else if (num == 2) return true;
            else if (num % 2 == 0) return false;

            double sqrtNum = Math.Sqrt(num);
            for (int i = 3; i <= sqrtNum; i += 2)
            {
                if (num % i == 0)
                {
                    return false;
                }
            }
            return true;
        }

        /// <summary>
        /// 順列mPnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Permutation(long m, long n)
        {
            long ret = 1;
            for (var i = 0; i < n; ++i)
            {
                ret *= m - i;
            }
            return ret;
        }
        /// <summary>
        /// 組み合わせmCnを計算する
        /// </summary>
        /// <param name="m"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static long Combination(long m, long n)
        {
            if (m == n || n == 0)
            {
                return 1;
            }
            long ans = 1;
            for (var i = 1; i <= n; ++i)
            {
                ans = ans * (m - i + 1) / i;
            }
            return ans;
        }

        static long Gcd(long m, long n)
        {
            if (n == 0) return m;
            return Gcd(n, m % n);
        }

        static long Lcm(long m, long n)
        {
            var gcd = Gcd(m, n);
            //return (m * n) / gcd;
            return m / gcd * n;//桁あふれ対策
        }
#endregion MyLib
    }
}