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 rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n)-1; i >= 0; i--)

using namespace std;
using ll = long long;
typedef pair<int, int> Pair;

const int MOD = 1000000007;

struct mint {
  int n;
  mint(int n_ = 0) : n(n_) {}
};
mint operator-(mint a) { return -a.n + MOD * (a.n != 0); }
mint operator+(mint a, mint b) {
  int x = a.n + b.n;
  return x - (x >= MOD) * MOD;
}
mint operator-(mint a, mint b) {
  int x = a.n - b.n;
  return x + (x < 0) * MOD;
}
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }
istream &operator>>(istream &i, mint &a) { return i >> a.n; }
ostream &operator<<(ostream &o, mint a) { return o << a.n; }

vector<mint> F_{1, 1}, R_{1, 1}, I_{0, 1};

void check_fact(int n) {
  for (int i = I_.size(); i <= n; i++) {
    I_.push_back(I_[MOD % i] * (MOD - MOD / i));
    F_.push_back(F_[i - 1] * i);
    R_.push_back(R_[i - 1] * I_[i]);
  }
}

mint I(int n) {
  check_fact(abs(n));
  return n >= 0 ? I_[n] : -I_[n];
}
mint F(int n) {
  check_fact(n);
  return n < 0 ? 0 : F_[n];
}
mint R(int n) {
  check_fact(n);
  return n < 0 ? 0 : R_[n];
}
mint C(int n, int r) { return F(n) * R(n - r) * R(r); }
mint P(int n, int r) { return F(n) * R(n - r); }
mint H(int n, int r) { return n == 0 ? (r == 0) : C(n + r - 1, r); }

int main() {
  string S;
  cin >> S;
  bool flag = true;

  for (int i = 0; i < S.length(); i++) {
    if ((i + 1) % 2 != 0) {
      if (!(S[i] == 'R' || S[i] == 'U' || S[i] == 'D'))  flag = false;
      } else {
        if (!(S[i] == 'L' || S[i] == 'U' || S[i] == 'D')) flag = false;
      }
    }
  cout << (flag ? "Yes" : "No") << endl;
}