Codeforces Round #778

D. Potion Brewing Class

/********************
Author:  Nanfeng1997
Contest: Codeforces - Codeforces Round #778 (Div. 1 + Div. 2, based on Technocup 2022 Final Round)
URL:     https://codeforces.com/contest/1654/problem/D
When:    2022-03-20 19:36:44

Memory:  256MB
Time:    3000ms
********************/
#include  <bits/stdc++.h>
#define _ 0
using namespace std;
using LL = long long;

template <typename T> void chkmax(T &x, T y) { x = max(x, y); }
template <typename T> void chkmin(T &x, T y) { x = min(x, y); }

const int MOD = 998244353;

inline int mod(int x) {return x >= MOD ? x - MOD : x;}

inline int ksm(int a, int b) {
  int ret = 1; a = mod(a);
  for(; b; b >>= 1, a = 1LL * a * a % MOD) if(b & 1) ret = 1LL * ret * a % MOD;
  return ret;
}

template<int MOD> 
struct modint {
  int x;
  modint() {x = 0; }
  modint(int y) {x = y;}
  inline modint inv() const { return modint{ksm(x, MOD - 2)}; }
  explicit inline operator int() { return x; }
  friend inline modint operator + (const modint &a, const modint& b) { return modint(mod(a.x + b.x)); }
  friend inline modint operator - (const modint &a, const modint& b) { return modint(mod(a.x - b.x + MOD)); }
  friend inline modint operator * (const modint &a, const modint& b) { return modint(1ll * a.x * b.x % MOD); }
  friend inline modint operator / (const modint &a, const modint& b) { return modint(1ll * a.x * b.inv().x % MOD); }
  friend inline modint operator - (const modint &a) { return modint(mod(MOD - a.x)); }
  friend inline modint& operator += (modint &a, const modint& b) { return a = a + b; }
  friend inline modint& operator -= (modint &a, const modint& b) { return a = a - b; }
  friend inline modint& operator *= (modint &a, const modint& b) { return a = a * b; }
  friend inline modint& operator /= (modint &a, const modint& b) { return a = a / b; }
  inline int operator == (const modint &b) { return x == b.x; }
  inline int operator != (const modint &b) { return x != b.x; }
  inline int operator < (const modint &a) { return x < a.x; }
  inline int operator <= (const modint &a) { return x <= a.x; }
  inline int operator > (const modint &a) { return x > a.x; }
  inline int operator >= (const modint &a) { return x >= a.x; }
};

typedef modint<MOD> mint;

inline mint ksm(mint a, int b) {
  mint ret = 1; 
  for(; b; b >>= 1, a = a * a ) if(b & 1) ret = ret * a ;
  return ret;
}


  
const int N = 2e5 + 10;
int d[N]; //d[i] 表示 i 的最小质因子是谁 
vector<int> factor[N]; //factor[i] 里面存储的是 i 的所有质因子(例如 factor[8] = {2,2,2})

inline void init() { //预处理
  for(int i = N - 1; i > 1; i -- ) {
    for(int j = i; j < N; j += i) {
      d[j] = i;
    }
  }
  for(int i = 1; i < N; i ++ ) {
    for(int j = i; j != 1; j /= d[j]) {
      factor[i].push_back(d[j]);
    }
  }
}

void solve() {
  int n; cin >> n;
  vector<vector<array<int, 3>> >son(n);

  set<int> s; //存储用到的质因子
  
  for(int i = 1; i < n; i ++ ) {
    int u, v, x, y; cin >> u >> v >> x >> y;
    -- u, -- v;
    son[u].push_back({v, x, y}); 
    son[v].push_back({u, y, x});
    //将用到的质因子存储到 set
    for(int &p: factor[x]) s.insert(p);
    for(int &p: factor[y]) s.insert(p);

  }

  vector<int> f(n + 1, 0), wf(n + 1, 0); 
  //f数组是用来每次枚举质因子是加还是减
  //wf 数组是用来记录在树的遍历的时候的每个质因子的最大值

  function<void(int, int)> dfs1 = [&] (int u, int fa) {
    for(auto &[v, x, y] : son[u]) {
      if(v == fa) continue;
      for(int &p: factor[y]) f[p] --;
      for(int &p: factor[x]) f[p] ++, chkmax(wf[p], f[p]);

      dfs1(v, u);

      for(int &p: factor[y]) f[p] ++;
      for(int &p: factor[x]) f[p] --;
    }
  };

  dfs1(0, -1); //进行树的遍历，求出 wf 数组

  mint ans = 0;

  function<void(int, int, mint)> dfs2 = [&] (int u, int fa, mint res) {
    ans += res;
    mint tem = res;//这里记录一下是因为每个分叉是互不影响的
    for(auto &[v, x, y]: son[u]) {
      if(v == fa) continue;
      res = tem;
      for(int &p: factor[y]) res *= p;  
      for(int &p: factor[x]) res /= p;
      //求出每个节点应该有的值
      dfs2(v, u, res);
    }

  };
  mint res = 1;
  //求出 1号节点应该有的值(即0号点)
  for(int x: s) {
    for(int i = 0; i < wf[x]; i ++ ) {
      res *= x;
    } 
  }

  dfs2(0, -1, res); //dfs2 求出答案!
  cout << (int)ans << "\n";

}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  init();
  int T = 1; cin >> T;
  while(T --) solve();

  return ~~(0 ^ _ ^ 0);
}