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 <stdio.h>
#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <limits>
#include <numeric>
#include <queue>
#include <set>
#include <bitset>
#include <map>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<long long> vll;
typedef queue<int, int> qii;
typedef pair<int, int> pii;
typedef pair<string, int> psi;

#define PI 3.14159265359
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repz(i,n) rep(i,0,n)

//int型を受け取ってその桁数を返す
int check_digits(int n) {
	int digits = 0;
	while (n != 0) {
		n /= 10;
		digits++;
	}
	return digits;
}

//int型を受け取ってその各桁の和を返す
int sum_digits(int n) {
	int sum=0;
	int N = n;
	while (N) {
		sum += N % 10;
		N /= 10;
	}
	return sum;
}


//最大公約数を返す関数
int gcd(int n, int m) {
	while (n != 0) {
		if (n < m) swap(n, m);
		n = n % m;
	}
	return m;
}


//2点間の距離を返す
float dis(int x1, int y1, int x2, int y2) {
	return sqrt(pow(abs(x1 - x2), 2) + pow(abs(y1 - y2), 2));
}


int main() {
	string s;
	cin >> s;
	repz(i, s.size()) {
		if (s[i] == 'L' && i % 2 == 0) { cout << "No" << endl; return 0; }
		if (s[i] == 'R' && i%2 == 1) { cout << "No" << endl; return 0; }
	}

	cout << "Yes" << endl;

	system("pause");
	return 0;
}