// #define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using ll = long long;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using vl = vector<ll>;
using vvl = vector<vl>;
using P = pair<ll,ll>;
#define pb push_back
#define int long long
#define double long double
#define INF (ll) 3e18
// Ctrl + Shift + B コンパイル
// Ctrl + C 中断
// ./m 実行


void solve(){
  int n, m; cin >> n >> m;
  vl a(n);
  vl b(m);
  rep(i,n) cin >> a[i];
  rep(i,m) cin >> b[i];
  if (n == 0 || m == 0){
    cout << "Yes" << endl;
    for(auto x : a) cout << "Red " << x << endl;
    for(auto x : b) cout << "Blue " << x << endl;
    return;
  }
  // abが共に等しいとき、移動することができる
  // a同士、b同士は相異なるため、同じ数値のペアだけを考えればよい
  set<int> amp;
  set<int> bmp;
  rep(i,n) amp.insert(a[i]);
  rep(i,m) bmp.insert(b[i]);
  vl ab;
  for(auto i : amp) if (bmp.count(i)) {
    ab.push_back(i);
  }
  if (ab.size() == 0) cout << "No" << endl;
  else {
    cout << "Yes" << endl;
    // 具体的な解答例を出力する
    for(auto i : amp) if (!bmp.count(i)) {
      cout << "Red " << i << endl;
    }
    cout << "Red " << ab[0] << endl;
    cout << "Blue " << ab[0] << endl;
    for(auto i : bmp) if (!amp.count(i)) {
      cout << "Blue " << i << endl;
    }
    bool c = 0;
    auto teban = [&]() -> string {
      c ^= 1;
      if (c) return "Blue ";
      if (!c) return "Red ";
    };
    for(int i = 1; i < ab.size(); ++i){
      cout << teban() << ab[i] << endl;
      cout << teban() << ab[i] << endl;
    }
  }
}

signed main(){
  int t; cin >> t;
  rep(_,t){
    solve();
  }
}