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)

#define DEBUG 0
/**
 * File    : B.cpp
 * Author  : Kazune Takahashi
 * Created : 6/4/2020, 3:17:31 PM
 * Powered by Visual Studio Code
 */
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <chrono>
#include <cmath>
#include <complex>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>
// ----- boost -----
#include <boost/rational.hpp>
#include <boost/multiprecision/cpp_int.hpp>
// ----- using directives and manipulations -----
using namespace std;
using boost::rational;
using boost::multiprecision::cpp_int;
using ll = long long;
using ld = long double;
template <typename T>
using max_heap = priority_queue<T>;
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
// ----- constexpr for Mint and Combination -----
constexpr ll MOD{1000000007LL};
// constexpr ll MOD{998244353LL}; // be careful
constexpr ll MAX_SIZE{3000010LL};
// constexpr ll MAX_SIZE{30000010LL}; // if 10^7 is needed
// ----- ch_max and ch_min -----
template <typename T>
void ch_max(T &left, T right)
{
  if (left < right)
  {
    left = right;
  }
}
template <typename T>
void ch_min(T &left, T right)
{
  if (left > right)
  {
    left = right;
  }
}
// ----- Mint -----
template <ll MOD = MOD>
class Mint
{
public:
  ll x;
  Mint() : x{0LL} {}
  Mint(ll x) : x{(x % MOD + MOD) % MOD} {}
  Mint operator-() const { return x ? MOD - x : 0; }
  Mint &operator+=(Mint const &a)
  {
    if ((x += a.x) >= MOD)
    {
      x -= MOD;
    }
    return *this;
  }
  Mint &operator-=(Mint const &a) { return *this += -a; }
  Mint &operator++() { return *this += 1; }
  Mint operator++(int)
  {
    Mint tmp{*this};
    ++*this;
    return tmp;
  }
  Mint &operator--() { return *this -= 1; }
  Mint operator--(int)
  {
    Mint tmp{*this};
    --*this;
    return tmp;
  }
  Mint &operator*=(Mint const &a)
  {
    (x *= a.x) %= MOD;
    return *this;
  }
  Mint &operator/=(Mint const &a)
  {
    Mint b{a};
    return *this *= b.power(MOD - 2);
  }
  Mint operator+(Mint const &a) const { return Mint(*this) += a; }
  Mint operator-(Mint const &a) const { return Mint(*this) -= a; }
  Mint operator*(Mint const &a) const { return Mint(*this) *= a; }
  Mint operator/(Mint const &a) const { return Mint(*this) /= a; }
  bool operator<(Mint const &a) const { return x < a.x; }
  bool operator<=(Mint const &a) const { return x <= a.x; }
  bool operator>(Mint const &a) const { return x > a.x; }
  bool operator>=(Mint const &a) const { return x >= a.x; }
  bool operator==(Mint const &a) const { return x == a.x; }
  bool operator!=(Mint const &a) const { return !(*this == a); }
  Mint power(ll N) const
  {
    if (N == 0)
    {
      return 1;
    }
    else if (N % 2 == 1)
    {
      return *this * power(N - 1);
    }
    else
    {
      Mint half = power(N / 2);
      return half * half;
    }
  }
};
template <ll MOD>
Mint<MOD> operator+(ll lhs, Mint<MOD> const &rhs)
{
  return rhs + lhs;
}
template <ll MOD>
Mint<MOD> operator-(ll lhs, Mint<MOD> const &rhs)
{
  return -rhs + lhs;
}
template <ll MOD>
Mint<MOD> operator*(ll lhs, Mint<MOD> const &rhs)
{
  return rhs * lhs;
}
template <ll MOD>
Mint<MOD> operator/(ll lhs, Mint<MOD> const &rhs)
{
  return Mint<MOD>{lhs} / rhs;
}
template <ll MOD>
istream &operator>>(istream &stream, Mint<MOD> &a)
{
  return stream >> a.x;
}
template <ll MOD>
ostream &operator<<(ostream &stream, Mint<MOD> const &a)
{
  return stream << a.x;
}
// ----- Combination -----
template <ll MOD = MOD, ll MAX_SIZE = MAX_SIZE>
class Combination
{
public:
  vector<Mint<MOD>> inv, fact, factinv;
  Combination() : inv(MAX_SIZE), fact(MAX_SIZE), factinv(MAX_SIZE)
  {
    inv[1] = 1;
    for (auto i = 2LL; i < MAX_SIZE; i++)
    {
      inv[i] = (-inv[MOD % i]) * (MOD / i);
    }
    fact[0] = factinv[0] = 1;
    for (auto i = 1LL; i < MAX_SIZE; i++)
    {
      fact[i] = Mint<MOD>(i) * fact[i - 1];
      factinv[i] = inv[i] * factinv[i - 1];
    }
  }
  Mint<MOD> operator()(int n, int k)
  {
    if (n >= 0 && k >= 0 && n - k >= 0)
    {
      return fact[n] * factinv[k] * factinv[n - k];
    }
    return 0;
  }
  Mint<MOD> catalan(int x, int y)
  {
    return (*this)(x + y, y) - (*this)(x + y, y - 1);
  }
};
// ----- for C++14 -----
using mint = Mint<MOD>;
using combination = Combination<MOD, MAX_SIZE>;
template <typename T>
T gcd(T x, T y) { return y ? gcd(y, x % y) : x; }
template <typename T>
T lcm(T x, T y) { return x / gcd(x, y) * y; }
// ----- for C++17 -----
template <typename T>
int popcount(T x) // C++20
{
  int ans{0};
  while (x != 0)
  {
    ans += x & 1;
    x >>= 1;
  }
  return ans;
}
// ----- frequently used constexpr -----
// constexpr double epsilon{1e-10};
// constexpr ll infty{1'000'000'000'000'010LL}; // or
// constexpr int infty{1'000'000'010};
// constexpr int dx[4] = {1, 0, -1, 0};
// constexpr int dy[4] = {0, 1, 0, -1};
// ----- Yes() and No() -----
void Yes()
{
  cout << "Yes" << endl;
  exit(0);
}
void No()
{
  cout << "No" << endl;
  exit(0);
}
// ----- main() -----

int main()
{
  string S;
  cin >> S;
  int N{static_cast<int>(S.size())};
  for (auto i = 0; i < N; ++i)
  {
    if ((i & 1) && S[i] == 'R')
    {
      No();
    }
    else if (!(i & 1) && S[i] == 'L')
    {
      No();
    }
  }
  Yes();
}