#include <iostream>
//前序遍历 递归的天花板 
using namespace std;
struct mobile{
	double lw, ll, rw, rl;
	double weight;
	mobile* left;
	mobile* right;

	mobile(double a, double b, double c, double d) : lw(a), ll(b), rw(c), rl(d), left(nullptr), right(nullptr) {}
};

bool solve(mobile* &x){
	double a, b, c, d;
	bool b1 = true, b2 = true;
	cin >> a >> b >> c >> d;
	x = new mobile(a, b, c, d);
	if(!x->lw) b1 = solve(x->left);
	if(!x->rw) b2 = solve(x->right);
	double leftWeight  = x->left  ? x->left->weight  : x->lw;
	double rightWeight = x->right ? x->right->weight : x->rw;
	x->weight = leftWeight + rightWeight;
	//关注return如果再solve一遍那就是多次递归了，不能再solve了 
	return b1 && b2 && (leftWeight * x->ll == rightWeight * x->rl);

}



int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    string tmp;
    getline(cin, tmp); // 读取第一行后的换行

    for (int i = 0; i < n; i++) {
        mobile* root = nullptr;
        bool ok = solve(root);
        cout << (ok ? "YES" : "NO") << "\n";

        if (i != n - 1) cout << "\n"; // 两个查询之间空一行
    }
	/*
	for(int i = 0; i < n - 1; i++){
		cin >> a >> b >> c >> d;
		mobile* tmp = new mobile(a, b, c, d);
		if(a && !b){
			
		}
	}
	这是不好的，应该是自然而然地形成一棵树 
	*/
	
	return 0;
}