Problem:
Score: 200 points

Problem Statement

Takahashi will do a tap dance. The dance is described by a string S where each character is L, R, U, or D. These characters indicate the positions on which Takahashi should step. He will follow these instructions one by one in order, starting with the first character.

S is said to be easily playable if and only if it satisfies both of the following conditions:

Every character in an odd position (1-st, 3-rd, 5-th, \ldots) is R, U, or D.

Every character in an even position (2-nd, 4-th, 6-th, \ldots) is L, U, or D.

Your task is to print Yes if S is easily playable, and No otherwise.

Constraints

S is a string of length between 1 and 100 (inclusive).

Each character of S is L, R, U, or D.

Input

Input is given from Standard Input in the following format:

S

Output

Print Yes if S is easily playable, and No otherwise.

Sample Input 1

RUDLUDR

Sample Output 1

Yes

Every character in an odd position (1-st, 3-rd, 5-th, 7-th) is R, U, or D.

Every character in an even position (2-nd, 4-th, 6-th) is L, U, or D.

Thus, S is easily playable.

Sample Input 2

DULL

Sample Output 2

No

The 3-rd character is not R, U, nor D, so S is not easily playable.

Sample Input 3

UUUUUUUUUUUUUUU

Sample Output 3

Yes

Sample Input 4

ULURU

Sample Output 4

No

Sample Input 5

RDULULDURURLRDULRLR

Sample Output 5

Yes

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        string read = rstr();
        string ans = "Yes";
        for(int i=0;i<read.Length;i++){
            if(read[i] == 'U' || read[i] == 'D'){
                continue;
            }
            if(i%2== 0 && read[i] == 'R'){
                continue;
            }
            if(i%2== 1 && read[i] == 'L'){
                continue;
            }
            ans = "No";
        }
        WriteLine(ans);

    
    }
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_double_array; }
    // A/B小数点以下切上
    static long divideup(long divided,long divid){ long temp_divide = divided/divid; if(divided % divid > 0){ temp_divide++; } return temp_divide; }
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
}
//知見
// (string) hogestr -> (char[]) hoge
    // char hoge = hogestr.ToCharArray();
// (char[]) hoge -> (string) hogestr
    // string hoge = new string(hoge);
// Hashtable ×  Dictionary ○
// Dictionary hoge<string,int>
    // hoge.ContainsKey(string) -> return boolean
    // hoge[string] -> return int
    // hoge[int] -> CE
// Dictionary hoge<int1,int2>
    // hoge<int1,int2>: hoge[int1] -> return int2
    // KeyValuePair<int,int> k <-> k.Key == int1 / k.Value == int2
// SortedDictionary() > SortedList() ???
// キュー　-> 幅優先 ???
// スタック -> 深さ優先 ???