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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

//char型のポインタ(=配列と意味は同じ)をString型と定義([0]から数える)
typedef char* String;

//S[0]から始まる文字列の長さを求める関数
int string_len(String str){
    int len=0;
    while(str[len]!=0){
        len++;
    }
    return len;
}

//文字列を標準入力から読み込み、それをString型のメモリを確保し直してくれる関数
//S[0]~S[n-1]までが入力で、S[n]は0になる
//文字列長さの上限は1000まで

String string_input(void){
    int i,len;
    char buf[1000];
    String str;
    scanf("%s",buf);//buf=入力した文字列が入るchar型の配列
    
    len=string_len(buf);//len=文字列の長さを表すint型変数
    NEW(str,len+1);//strはただのポインタだから、ここでNEWを使ってメモリを確保しなくちゃいけない
    for(i=0;i<len;i++){
        str[i]=buf[i];
    }
    str[len]=0;
    
    return str;
}

int main(void){
    String S=string_input(); //入力文字列

    int N=string_len(S);

    int count; //Yesなら1,Noなら0にする
    for(int i=0;i<N;i++){
        if(i%2==0){
            if(S[i]=='L'){
                count=0;
                break;
            }
            else{
                count=1;
            }
        }
        else{
            if(S[i]=='R'){
                count=0;
                break;
            }
            else{
                count=1;
            }
        }
    }

    if(count==1){
        printf("Yes\n");
    }
    else{
        printf("No\n");
    }
    return 0;
}