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 (Python):
S = input()
for i in range(len(S)):
  if S[i] == "U":
    continue
  if S[i] == "D":
    continue
  if i % 2 == 1:
    if S[i] == "L":
      continue
  else:
    if S[i] == "R":
      continue
  print("No")
  exit()
print("Yes")

Solution (Python):

import sys
sys.setrecursionlimit(300000)

YES = "Yes"  
NO = "No"  

def solve(S: str):
    ret = YES
    for i in range(len(S)):
        if i % 2 == 0 and S[i] in ['R', 'U', 'D']:
            continue
        if i % 2 == 1 and S[i] in ['L', 'U', 'D']:
            continue
        ret = NO
        break
    print(ret)
    return


def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    S = next(tokens)  
    solve(S)

if __name__ == '__main__':
    main()

Solution (Python):
import sys,collections as cl,bisect as bs
sys.setrecursionlimit(100000)
mod = 10**9+7
Max = sys.maxsize
def l(): 
    return list(map(int,input().split()))
def m(): 
    return map(int,input().split())
def onem(): 
    return int(input())
def s(x): 
    a = []
    aa = x[0]
    su = 1
    for i in range(len(x)-1):
        if aa != x[i+1]:
            a.append([aa,su])
            aa = x[i+1]
            su = 1
        else:
            su += 1
    a.append([aa,su])
    return a
def jo(x): 
    return " ".join(map(str,x))
def max2(x): 
    return max(map(max,x))
def In(x,a): 
    k = bs.bisect_left(a,x)
    if k != len(a) and a[k] ==  x:
        return True
    else:
        return False

s = list(input())
co = 0
for i in range(len(s)):
    if i % 2 == 0:
        if s[i] == "R" or s[i] == "U" or s[i] == "D":
            co += 1
        else:
            print("No")
            exit()
    else:
        if s[i] == "L" or s[i] == "U" or s[i] == "D":
            co += 1
        else:
            print("No")
            exit()
print("Yes")

Solution (Python):
import sys
p = input()
for i in range(len(p)):
    if i % 2 == 0:
        if p[i] == "L":
            print("No")
            sys.exit()
    else:
        if p[i] == "R":
            print("No")
            sys.exit()
print("Yes")

Solution (Python):
import math
import heapq
from bisect import bisect_right, bisect_left
from collections import defaultdict, Counter
import itertools
import sys
sys.setrecursionlimit(10**7)
MOD = 10**9 + 7
INF = float("inf")


def main():
    S = list(input())
    flg = True
    for i in range(len(S)):
        if i % 2 == 0:
            if S[i] == "R" or S[i] == "U" or S[i] == "D":
                pass
            else:
                print("No")
                exit()
        else:
            if S[i] == "L" or S[i] == "U" or S[i] == "D":
                pass
            else:
                print("No")
                exit()
    print("Yes")


if __name__ == '__main__':
    main()

Solution (Python):
s=input()
n=len(s)
j=True
for i in range(0,n,2):
	if s[i] not in list("RUD"):
		j=False
for i in range(1,n,2):
	if s[i] not in list("LUD"):
		j=False
print("Yes" if j else "No")

Solution (Python):
s = input()
n = len(s)
ok = True
for i in range(0, n, 2):
	if s[i] == "L":
		ok = False
for i in range(1, n, 2):
	if s[i] == "R":
		ok = False
if ok:
	print("Yes")
else:
	print("No")

Solution (Python):
s = list(input())
odd = ['R', 'U', 'D']
even = ['L', 'U', 'D']

for i in range(len(s)):
    if i % 2 == 0:
        if s[i] not in odd:
            print('No')
            exit()
    else:
        if s[i] not in even:
            print('No')
            exit()
print('Yes')

Solution (Python):




def solve():
  S = str(input())
  for i,c in enumerate(S):
    if i%2==0:
      if c not in "RUD":
        return False
    else:
      if c not in "LUD":
        return False
  return True

if solve():
  print("Yes")
else:
  print("No")

Solution (Python):


from collections import deque, defaultdict
import itertools as ite
import sys
import math
import decimal

sys.setrecursionlimit(1000000)

INF = 10 ** 18
MOD = 10 ** 9 + 7

S = raw_input()
flag = True
for i in range(len(S)):
    if i % 2 == 0:
        if not S[i] in "RUD":
            flag = False
    else:
        if not S[i] in "LUD":
            flag = False
print "Yes" if flag else "No"

Solution (Python):

from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007


def A():

    return


def B():
    s = S()
    for i in range(len(s)):
        if i%2:
            if s[i] not in ["L","U","D"]:
                print("No")
                return
        else:
            if s[i] not in ["R","U","D"]:
                print("No")
                return
    print("Yes")
    return


def C():

    return


def D():

    return


def E():

    return


def F():

    return


if __name__ == "__main__":
    B()