#include<cstdio>

#include<cstring>

#include<iostream>

#include<string>

#include<algorithm>

using namespace std;

struct Node {

        Node() {

                wl=wr=dl=dr=0;

                l=r=0;

        }

        int wl;

        int dl;

        int wr;

        int dr;

        Node* l;

        Node* r;

};

Node* build()

{

        int wl, wr, dl, dr;

        scanf("%d%d%d%d", &wl, &dl, &wr, &dr);

        Node* root=new Node;

        if(wl==0)

        {

                Node* left=build();

                wl=left->wl + left->wr;

                root->l=left;

        }

        if(wr==0)

        {

                Node* right=build();

                wr=right->wl + right->wr;

                root->r=right;

        }

        root->wl=wl;

        root->wr=wr;

        root->dl=dl;

        root->dr=dr;

        return root;

}

bool equilibrium(Node* root)

{

        if(!root)

                return true;

        bool el=equilibrium(root->l);

        bool er=equilibrium(root->r);

        if(el&&er)

        {

                return (root->wl * root->dl == root->wr * root->dr);

        }

        else

        {

                return false;

        }

}

int main()

{

#ifndef ONLINE_JUDGE

        freopen("./uva839.in", "r", stdin);

#endif

        int T;

        scanf("%d", &T);

        while(T--) {

                Node* root=build();

                if(equilibrium(root))

                        printf("YES\n");

                else

                        printf("NO\n");

                if(T!=0)

                        printf("\n");

        }

    return 0;

}

// UVa839 Not so Mobile

// Rujia Liu

// 题意：输入一个树状天平，根据力矩相等原则判断是否平衡。采用递归方式输入，0表示中间结点

// 算法：在“建树”时直接读入并判断，并且无须把树保存下来

#include<iostream>

using namespace std;

// 输入一个子天平，返回子天平是否平衡，参数W修改为子天平的总重量

bool solve(int& W) {

  int W1, D1, W2, D2;

  bool b1 = true, b2 = true;

  cin >> W1 >> D1 >> W2 >> D2;

  if(!W1) b1 = solve(W1);

  if(!W2) b2 = solve(W2);

  W = W1 + W2;

  return b1 && b2 && (W1 * D1 == W2 * D2);

}

int main() {

  int T, W;

  cin >> T;

  while(T--) {

    if(solve(W)) cout << "YES\n"; else cout << "NO\n";

    if(T) cout << "\n";

  }

  return 0;

}