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 <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
using pii=pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
#define all(c) begin(c),end(c)
#define rall(c) rbegin(c),rend(c)
#define fore(x,c) for(auto &&x:c)
#define rep(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define rrep(i, a, n) for(int i=(int)(n-1);i>=a;--i)
#define sz(c) ((int)c.size())
#define contains(c,x) (c.find(x)!=end(c))
#define inseg(l,x,r) ((l)<=(x)&&(x)<(r))
#define dump(...)
#define pb push_back
#define _ 0
const signed INF_=1001001001; const long long INF=1001001001001001001LL;
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    for (auto i = begin(v); i != end(v); i++) os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    for (auto i = begin(v); i != end(v); i++) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T, class U> bool chmax(T &a,const U &b){return a<b?a=b,1:0;}
template<class T, class U> bool chmin(T &a,const U &b){return a>b?a=b,1:0;}
template <class T> void psum(T& c) {partial_sum(begin(c), end(c), begin(c));}
template<class T> using heap=priority_queue<T,vector<T>,greater<T>>;
struct before_main_function {
    before_main_function() {
        cin.tie(nullptr); ios::sync_with_stdio(0);
        cout << setprecision(15) << fixed;
        // #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    set<char> L = {'L', 'U', 'D'};
    set<char> R = {'R', 'U', 'D'};
    string s;
    cin >> s;
    bool ng = false;
    rep(i, 0, sz(s)) {
        if (i % 2) {
            if (!contains(L, s[i])) ng = true;
        } else {
            if (!contains(R, s[i])) ng = true;
        }
    }
    cout << (ng ? "No" : "Yes") << endl;
    return (0^_^0);
}