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)

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define                           int                         long long
#define                           fi                          first
#define                           si                          second
#define                           mp                          make_pair
#define                           pb                          push_back
#define                           pi                          pair<int,int>
#define                           f(i,l,r)                    for(int i=l;i<=r;i++)
#define                           rf(i,r,l)                   for(int i=r;i>=l;i--)
#define                           done(i)                     cout<<"done = "<<i<<endl;
#define                           fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);




using namespace std;

const int inf=1e18;
const int mod=1e9+7;
const int M=100009;
inline int bigmod(int B,int P){int R=1;while(P>0){if(P&1){R=(R*B)%mod;}P>>=1;B=(B*B)%mod;}return R;}


 main()

{
    fast
    string s;
    cin>>s;
    int n=s.size();
    s="#"+s;
    for(int i=1;i<=n;i++)
    {
       if(i%2==1)
       {if(!(s[i]=='R' || s[i]=='U' || s[i]=='D'))
        {
            cout<<"No"<<endl;
            return 0;
        }

       }
        else
       {if(!(s[i]=='L' || s[i]=='U' || s[i]=='D'))
        {
            cout<<"No"<<endl;
            return 0;
        }

       }
    }
    cout<<"Yes"<<endl;


















    return 0;

}