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>
#include <string.h>
#include <math.h>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

long long  factorial(long long n) {
    if (n > 0) {
        return n * factorial(n - 1);
    } else {
        return 1;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}

//int ()



int main()
{
  int a ,b, c, d, e, X, Y;
//  long long  a, b, c, d, e, X, Y;
  //  double  c;
//  int  A[200003]={};
  int  A[100003]={};
  int  B[100003]={};
  int  C[100003][11]={};

  int  flg=0;

//  long long  a, b, c, d, X, Y;
  char s[101][101]={};
  char t[101]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p,q;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int len=0;
  int l=0;
  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  int ans=0;
  int max=0;
  int sum=0;
  int addr=0;
  int ptn=0;
  long long  tmp=0;
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%s", &t);
  len = strlen(t);
  for(i=0;i<len;i++){
    if(i%2==1){
      if( !( (t[i]=='L') || (t[i]=='U') ||  (t[i]=='D'))){
        printf("No\n");
        return 0;
      }
    }
    else {
      if( !( (t[i]=='R') || (t[i]=='U') ||  (t[i]=='D'))){
        printf("No\n");
        return 0;
      }
    }
  }
         
  printf("Yes\n");     
 

    
//    printf("%s %d %d\n", s[i],C[i][0],C[i][1]);
   // printf("%d\n", C[i][1]);
/*  for(i=0;i<addr;i++){
//  scanf("%lld%lld", &B[0][i],&B[1][i]);
//    ans=ans+A[i];
    printf("%lld ", A[i]);
  }
*/
//    printf("%d\n", cnt);
   

return 0;
}