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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#include<cmath>


#define REP(i,n) for (int i = 0; (i) < (n); ++ (i))
#define FOR(i,n) for (int i = 1; (i) <= (n); ++ (i))
#define dump(x)  cout << #x << " = " << (x) << endl;
#define pb push_back
#define int long long
//#define lint long long

const int INF = 1e18;
const int MOD = 1e9+7;
//const lint LINF = 1e18;
const double eps = 0.000000001;//もとの値の10^(-16)まで

using namespace std;

//typedef pair<int, int> P;
//priority_queue< P, vector<P>, greater<P> > q;//ダイクストラの時、greaterで小さい順

signed main(){
  string s;
  cin  >> s;
  string ans = "Yes";
  REP(i,s.size()){
    if(i%2==0){
      if(s[i]=='L') ans = "No";
    }
    else{
      if(s[i]=='R') ans = "No";
    }
  }
  cout << ans << endl;
  return 0;
}