description: 'Twas the night before Christmas, and Santa's frantically setting up his new Christmas tree! There are n nodes in the tree, connected by n-1 edges. On each edge of the tree, there's a set of Christmas lights, which can be represented by an integer in binary representation.
  
  <image>
  
  He has m elves come over and admire his tree. Each elf is assigned two nodes, a and b, and that elf looks at all lights on the simple path between the two nodes. After this, the elf's favorite number becomes the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of the values of the lights on the edges in that path.
  
  However, the North Pole has been recovering from a nasty bout of flu. Because of this, Santa forgot some of the configurations of lights he had put on the tree, and he has already left the North Pole! Fortunately, the elves came to the rescue, and each one told Santa what pair of nodes he was assigned (a_i, b_i), as well as the parity of the number of set bits in his favorite number. In other words, he remembers whether the number of 1's when his favorite number is written in binary is odd or even.
  
  Help Santa determine if it's possible that the memories are consistent, and if it is, remember what his tree looked like, and maybe you'll go down in history!
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains two integers, n and m (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the size of tree and the number of elves respectively.
  
  The next n-1 lines of each test case each contains three integers, x, y, and v (1 ≤ x, y ≤ n; -1 ≤ v < 2^{30}) — meaning that there's an edge between nodes x and y. If 
  
    * v = -1: Santa doesn't remember what the set of lights were on for this edge. 
    * v ≥ 0: The set of lights on the edge is v. 
  
  
  
  The next m lines of each test case each contains three integers, a, b, and p (1 ≤ a, b ≤ n; a ≠ b; 0 ≤ p ≤ 1) — the nodes that the elf was assigned to, and the parity of the number of set bits in the elf's favorite number.
  
  It is guaranteed that the sum of all n and the sum of all m don't exceed 2 ⋅ 10^5 each.
  
  It is guaranteed that the given edges form a tree.
  
  Output
  
  For each test case, first print either YES or NO (in any case), whether there's a tree consistent with Santa's memory or not. 
  
  If the answer is YES, print n-1 lines each containing three integers: x, y, and v (1 ≤ x, y ≤ n; 0 ≤ v < 2^{30}) — the edge and the integer on that edge. The set of edges must be the same as in the input, and if the value of some edge was specified earlier, it can not change. You can print the edges in any order.
  
  If there are multiple answers, print any.
  
  Example
  
  Input
  
  
  4
  6 5
  1 2 -1
  1 3 1
  4 2 7
  6 3 0
  2 5 -1
  2 3 1
  2 5 0
  5 6 1
  6 1 1
  4 5 1
  5 3
  1 2 -1
  1 3 -1
  1 4 1
  4 5 -1
  2 4 0
  3 4 1
  2 3 1
  3 3
  1 2 -1
  1 3 -1
  1 2 0
  1 3 1
  2 3 0
  2 1
  1 2 1
  1 2 0
  
  
  Output
  
  
  YES
  1 2 0
  1 3 1
  2 4 7
  3 6 0
  2 5 0
  YES
  1 2 1
  1 3 0
  1 4 1
  4 5 1
  NO
  NO
  
  Note
  
  The first test case is the image in the statement.
  
  One possible answer is assigning the value of the edge (1, 2) to 5, and the value of the edge (2, 5) to 3. This is correct because: 
  
    * The first elf goes from node 2 to node 3. This elf's favorite number is 4, so he remembers the value 1 (as 4 has an odd number of 1 bits in its binary representation). 
    * The second elf goes from node 2 to node 5. This elf's favorite number is 3, so he remembers the value 0 (as 3 has an even number of 1 bits in its binary representation). 
    * The third elf goes from node 5 to node 6. This elf's favorite number is 7, so he remembers the value 1 (as 7 has an odd number of 1 bits in its binary representation). 
    * The fourth elf goes from node 6 to node 1. This elf's favorite number is 1, so he remembers the value 1 (as 1 has an odd number of 1 bits in its binary representation). 
    * The fifth elf goes from node 4 to node 5. This elf's favorite number is 4, so he remembers the number 1 (as 4 has an odd number of 1 bits in its binary representation). 
  
  
  
  Note that there are other possible answers.
solution: #include <bits/stdc++.h>
  const double PI = 3.1415926535897932384626433;
  using namespace std;
  struct edge {
    long long to, cost;
    edge() {}
    edge(long long a, long long b) { to = a, cost = b; }
  };
  const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
  const int mod = 1000000007;
  struct mint {
    int x;
    mint(long long y = 0) {
      if (y < 0 || y >= mod) y = (y % mod + mod) % mod;
      x = y;
    }
    mint(const mint &ope) { x = ope.x; }
    mint operator-() { return mint(-x); }
    mint operator+(const mint &ope) { return mint(x) += ope; }
    mint operator-(const mint &ope) { return mint(x) -= ope; }
    mint operator*(const mint &ope) { return mint(x) *= ope; }
    mint operator/(const mint &ope) { return mint(x) /= ope; }
    mint &operator+=(const mint &ope) {
      x += ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint &operator-=(const mint &ope) {
      x += mod - ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint &operator*=(const mint &ope) {
      long long tmp = x;
      tmp *= ope.x, tmp %= mod;
      x = tmp;
      return *this;
    }
    mint &operator/=(const mint &ope) {
      long long n = mod - 2;
      mint mul = ope;
      while (n) {
        if (n & 1) *this *= mul;
        mul *= mul;
        n >>= 1;
      }
      return *this;
    }
    mint inverse() { return mint(1) / *this; }
    bool operator==(const mint &ope) { return x == ope.x; }
    bool operator!=(const mint &ope) { return x != ope.x; }
    bool operator<(const mint &ope) const { return x < ope.x; }
  };
  mint modpow(mint a, long long n) {
    if (n == 0) return mint(1);
    if (n % 2)
      return a * modpow(a, n - 1);
    else
      return modpow(a * a, n / 2);
  }
  istream &operator>>(istream &is, mint &ope) {
    long long t;
    is >> t, ope.x = t;
    return is;
  }
  ostream &operator<<(ostream &os, mint &ope) { return os << ope.x; }
  ostream &operator<<(ostream &os, const mint &ope) { return os << ope.x; }
  long long modpow(long long a, long long n, long long mod) {
    if (n == 0) return 1;
    if (n % 2)
      return ((a % mod) * (modpow(a, n - 1, mod) % mod)) % mod;
    else
      return modpow((a * a) % mod, n / 2, mod) % mod;
  }
  vector<mint> fact, fact_inv;
  void make_fact(int n) {
    fact.resize(n + 1), fact_inv.resize(n + 1);
    fact[0] = mint(1);
    for (long long i = (1); (i) <= (n); (i)++) fact[i] = fact[i - 1] * mint(i);
    fact_inv[n] = fact[n].inverse();
    for (long long i = (n - 1); (i) >= (0); (i)--)
      fact_inv[i] = fact_inv[i + 1] * mint(i + 1);
  }
  mint comb(int n, int k) {
    if (n < 0 || k < 0 || n < k) return mint(0);
    return fact[n] * fact_inv[k] * fact_inv[n - k];
  }
  mint perm(int n, int k) { return comb(n, k) * fact[k]; }
  vector<int> prime, pvec, qrime;
  void make_prime(int n) {
    prime.resize(n + 1);
    for (long long i = (2); (i) <= (n); (i)++) {
      if (prime[i] == 0) pvec.push_back(i), prime[i] = i;
      for (auto p : pvec) {
        if (i * p > n || p > prime[i]) break;
        prime[i * p] = p;
      }
    }
  }
  void make_qrime(int n) {
    qrime.resize(n + 1);
    for (long long i = (2); (i) <= (n); (i)++) {
      int ni = i / prime[i];
      if (prime[i] == prime[ni])
        qrime[i] = qrime[ni] * prime[i];
      else
        qrime[i] = prime[i];
    }
  }
  bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
  void mark() { cout << "*" << endl; }
  void yes() { cout << "YES" << endl; }
  void no() { cout << "NO" << endl; }
  long long floor(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return a / b;
    else
      return -((-a + b - 1) / b);
  }
  long long ceil(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return (a + b - 1) / b;
    else
      return -((-a) / b);
  }
  long long modulo(long long a, long long b) {
    b = abs(b);
    return a - floor(a, b) * b;
  }
  long long sgn(long long x) {
    if (x > 0) return 1;
    if (x < 0) return -1;
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long arith(long long x) { return x * (x + 1) / 2; }
  long long arith(long long l, long long r) { return arith(r) - arith(l - 1); }
  long long digitnum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret++;
    return ret;
  }
  long long digitsum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret += x % b;
    return ret;
  }
  string lltos(long long x) {
    string ret;
    for (; x; x /= 10) ret += x % 10 + '0';
    reverse((ret).begin(), (ret).end());
    return ret;
  }
  long long stoll(string &s) {
    long long ret = 0;
    for (auto c : s) ret *= 10, ret += c - '0';
    return ret;
  }
  template <typename T>
  void uniq(T &vec) {
    sort((vec).begin(), (vec).end());
    vec.erase(unique((vec).begin(), (vec).end()), vec.end());
  }
  int popcount(unsigned long long x) {
    x -= ((x >> 1) & 0x5555555555555555ULL),
        x = (x & 0x3333333333333333ULL) + ((x >> 2) & 0x3333333333333333ULL);
    return (((x + (x >> 4)) & 0x0F0F0F0F0F0F0F0FULL) * 0x0101010101010101ULL) >>
           56;
  }
  template <class S, class T>
  pair<S, T> &operator+=(pair<S, T> &s, const pair<S, T> &t) {
    s.first += t.first, s.second += t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T> &operator-=(pair<S, T> &s, const pair<S, T> &t) {
    s.first -= t.first, s.second -= t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first + t.first, s.second + t.second);
  }
  template <class S, class T>
  pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first - t.first, s.second - t.second);
  }
  template <class T>
  T dot(const pair<T, T> &s, const pair<T, T> &t) {
    return s.first * t.first + s.second * t.second;
  }
  template <class T>
  T cross(const pair<T, T> &s, const pair<T, T> &t) {
    return s.first * t.second - s.second * t.first;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> &vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, deque<T> &deq) {
    for (long long i = 0; (i) < (long long)(deq).size(); (i)++)
      os << deq[i] << " ";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, pair<T, U> &ope) {
    os << "(" << ope.first << ", " << ope.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, const pair<T, U> &ope) {
    os << "(" << ope.first << ", " << ope.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, map<T, U> &ope) {
    for (auto p : ope) os << "(" << p.first << ", " << p.second << "),";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> &ope) {
    for (auto x : ope) os << x << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> &ope) {
    for (auto x : ope) os << x << " ";
    return os;
  }
  template <typename T>
  void outa(T a[], long long s, long long t) {
    for (long long i = (s); (i) <= (t); (i)++) {
      cout << a[i];
      if (i < t) cout << " ";
    }
    cout << endl;
  }
  void dump_func() { cout << endl; }
  template <class Head, class... Tail>
  void dump_func(Head &&head, Tail &&...tail) {
    cout << head;
    if (sizeof...(Tail) > 0) cout << " ";
    dump_func(std::move(tail)...);
  }
  struct SCC {
    int n;
    vector<vector<int> > G, revG, compG;
    vector<int> used, scc, topo;
    int sccid, sccnum;
    void tpdfs(int v) {
      used[v] = 1;
      for (int i = 0; i < G[v].size(); i++) {
        if (!used[G[v][i]]) tpdfs(G[v][i]);
      }
      topo.push_back(v);
    }
    void sccdfs(int v, int id) {
      used[v] = 1;
      scc[v] = id;
      for (int i = 0; i < revG[v].size(); i++) {
        if (!used[revG[v][i]]) sccdfs(revG[v][i], id);
      }
    }
    SCC() {}
    SCC(int n) {
      this->n = n;
      G.resize(n + 1);
      revG.resize(n + 1);
      used.resize(n + 1);
    }
    void init() {
      for (int i = 1; i <= n; i++) {
        G[i].clear(), revG[i].clear();
        used[i] = 0;
      }
      topo.clear();
    }
    void add_edge(int u, int v) { G[u].push_back(v); }
    void tpsort() {
      topo.clear();
      for (int i = 1; i <= n; i++) used[i] = 0;
      for (int i = 1; i <= n; i++)
        if (!used[i]) tpdfs(i);
      reverse(topo.begin(), topo.end());
    }
    bool checkDAG() {
      for (int i = 1; i <= n; i++) used[i] = 0;
      for (int i = 0; i < topo.size(); i++) {
        int v = topo[i];
        used[v] = 1;
        for (int j = 0; j < G[v].size(); j++) {
          if (used[G[v][j]]) return false;
        }
      }
      return true;
    }
    int calcSCC() {
      scc.resize(n + 1);
      for (int i = 1; i <= n; i++) revG[i].clear();
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < G[i].size(); j++) {
          revG[G[i][j]].push_back(i);
        }
      }
      sccid = 1;
      for (int i = 1; i <= n; i++) used[i] = 0;
      for (int i = 0; i < topo.size(); i++)
        if (!used[topo[i]]) sccdfs(topo[i], sccid++);
      return sccnum = sccid - 1;
    }
    void compressSCC(bool simple = false) {
      compG.resize(sccnum + 1);
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < G[i].size(); j++) {
          int u = G[i][j];
          if (scc[i] != scc[u]) compG[scc[i]].push_back(scc[u]);
        }
      }
      if (simple) {
        for (int i = 1; i <= sccnum; i++) {
          sort(compG[i].begin(), compG[i].end());
          compG[i].erase(unique(compG[i].begin(), compG[i].end()),
                         compG[i].end());
        }
      }
    }
  };
  struct TwoSat {
    int n;
    SCC scc;
    vector<int> assign;
    TwoSat() {}
    TwoSat(int n) {
      this->n = n;
      scc = SCC(2 * n + 1);
    }
    void addClause(int a, int b) {
      scc.add_edge(-a + n + 1, b + n + 1);
      scc.add_edge(-b + n + 1, a + n + 1);
    }
    bool calc() {
      scc.tpsort(), scc.calcSCC();
      assign.resize(n + 1);
      for (int i = 1; i <= n; i++) {
        if (scc.scc[i + n + 1] == scc.scc[-i + n + 1]) return false;
        if (scc.scc[i + n + 1] > scc.scc[-i + n + 1])
          assign[i] = 1;
        else
          assign[i] = -1;
      }
      return true;
    }
  };
  long long T;
  long long n, m;
  TwoSat ts;
  long long w[200005];
  vector<edge> G[200005];
  void dfs(int v, int p) {
    for (auto e : G[v]) {
      long long u = e.to, id = e.cost;
      if (u == p) continue;
      if (w[id] == -1) {
        if (ts.assign[u] == ts.assign[v])
          w[id] = 0;
        else
          w[id] = 1;
      }
      cout << u << " " << v << " " << w[id] << "\n";
      dfs(u, v);
    }
  }
  int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> T;
    while (T--) {
      cin >> n >> m;
      ts = TwoSat(n);
      for (long long i = (1); (i) <= (n); (i)++) G[i].clear();
      long long u, v;
      for (long long i = (1); (i) <= (n - 1); (i)++) {
        cin >> u >> v >> w[i];
        G[u].push_back(edge(v, i));
        G[v].push_back(edge(u, i));
        if (w[i] != -1) {
          if (popcount(w[i]) % 2 == 0)
            ts.addClause(-u, v), ts.addClause(-v, u);
          else
            ts.addClause(-u, -v), ts.addClause(v, u);
        }
      }
      long long p;
      for (long long i = (1); (i) <= (m); (i)++) {
        cin >> u >> v >> p;
        if (p == 0)
          ts.addClause(-u, v), ts.addClause(-v, u);
        else
          ts.addClause(-u, -v), ts.addClause(v, u);
      }
      if (!ts.calc()) {
        cout << "NO"
             << "\n";
        continue;
      }
      cout << "YES"
           << "\n";
      dfs(1, -1);
    }
    return 0;
  }
  