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>
#define int         int64_t
#define pb          push_back
#define endl        '\n'
#define pll         pair<int,int>
#define vll          vector<int>
#define all(a)      (a).begin(),(a).end()
#define x           first
#define y           second
#define sz(x)       (int)x.size()
#define hell        1000000007
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mi        map<int,int>
#define gcd(a,b)    __gcd((a),(b))
#define lcm(a,b)    ((a)*(b)) / gcd((a),(b))
#define ios     ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); 
int n,i,j,k,l,sum=0,flag=0,t,a[200005],ans=0;
void solve(){
    string s;
    cin>>s;
    for(int i=0;i<s.size();i++){
        if((i%2==0 && s[i]=='L')||(i%2==1 && s[i]=='R')){
            cout<<"No\n";
            return;
        }
    }
    cout<<"Yes\n";
}
signed main()
{
    ios
    int test=1;
    //cin>>test;
    while(test--){
        solve();
    }
    return 0;
}