#include <algorithm>
#include <atcoder/all>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#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 <utility>
#include <vector>
using namespace std;
using namespace atcoder;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n) - 1; i >= 0; i--)
#define repk(i, k, n) for (int i = k; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define mod1 1000000007
#define mod2 998244353
#define mod3 100000007
#define vi vector<int>
#define vs vector<string>
#define vc vector<char>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vvl vector<vector<ll>>
#define vvb vector<vector<bool>>
#define vvvi vector<vector<vector<int>>>
#define vvvl vector<vector<vector<ll>>>
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define vpii vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define vvpii vector<vector<pair<int, int>>>
#define vvpll vector<vector<pair<ll, ll>>>

template <typename T>
void debug(T e) {
    cerr << e << endl;
}

template <typename T>
void debug(vector<T> &v) {
    rep(i, v.size()) { cerr << v[i] << " "; }
    cerr << endl;
}

template <typename T>
void debug(vector<vector<T>> &v) {
    rep(i, v.size()) {
        rep(j, v[i].size()) { cerr << v[i][j] << " "; }
        cerr << endl;
    }
}

template <typename T>
void debug(vector<pair<T, T>> &v) {
    rep(i, v.size()) { cerr << v[i].first << " " << v[i].second << endl; }
}

template <typename T>
void debug(set<T> &st) {
    for (auto itr = st.begin(); itr != st.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(multiset<T> &ms) {
    for (auto itr = ms.begin(); itr != ms.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(map<T, T> &mp) {
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        cerr << itr->first << " " << itr->second << endl;
    }
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << H << " ";
    debug_out(T...);
}

using mint = modint998244353;

void debug_mint1(vector<mint> &vec) {
    for (int i = 0; i < vec.size(); i++) {
        cerr << vec[i].val() << " ";
    }
    cerr << endl;
}

void debug_mint2(vector<vector<mint>> &vec) {
    for (int i = 0; i < vec.size(); i++) {
        for (int j = 0; j < vec[i].size(); j++) {
            cerr << vec[i][j].val() << " ";
        }
        cerr << endl;
    }
}

int main() {
    ll N;
    cin >> N;
    string S;
    cin >> S;

    if (N == 1){
        if (S == "o") cout << "No" << endl;
        else{
            cout << "Yes" << endl;
            cout << "L" << endl;
        }
        return 0;
    }
    // 最初の 1 手を両方試して、あとは芋づる式に
    
    string a1 = "";
    string a2 = "";

    vector<bool> has(N - 1, false);
    rep(i, N - 1){
        if (S[i] != S[i + 1]) has[i] = true;
    }
    a1 += 'L';
    if (has[0]) has[0] = false;
    else has[0] = true;
    for (ll i = 1; i < N; i++){
        if (has[i - 1]){
            has[i - 1] = false;
            a1 += 'R';
        }
        else{
            if (i < N - 1){
                if (has[i]) has[i] = false;
                else has[i] = true;
            }
            a1 += 'L';
        }
    }

    // 実際に操作をしてみて、S_0 が o なら出力
    ll times = 0;
    rep(i, N){
        if (a1[i] == 'L') times++;
    }
    if ((times % 2 == 0 && S[0] == 'o') || (times % 2 == 1 && S[0] == 'x')){
        cout << "Yes" << endl;
        cout << a1 << endl;
        return 0;
    }

    rep(i, N - 1){
        if (S[i] != S[i + 1]) has[i] = true;
        else has[i] = false;
    }
    a2 += 'R';
    for (ll i = 1; i < N; i++){
        if (has[i - 1]){
            has[i - 1] = false;
            a2 += 'R';
        }
        else{
            if (i < N - 1){
                if (has[i]) has[i] = false;
                else has[i] = true;
            }
            a2 += 'L';
        }
    }

    times = 0;
    rep(i, N){
        if (a2[i] == 'R') times++;
    }
    if ((times % 2 == 0 && S[N - 1] == 'o') || (times % 2 == 1 && S[N - 1] == 'x')){
        cout << "Yes" << endl;
        cout << a2 << endl;
        return 0;
    }

    cout << "No" << endl;
}