#include<bits/stdc++.h>
using namespace std;
#define int long long 
#define PII pair<int,int>
const int N=5e5+10;
const int mod=998244353;
int n,m;
int k;
// vector<PII>v;
void solve()
{
    
    cin>>n;
    int a,b;
    vector<array<int,2>>p(n);
    for(int i=0;i<n;i++)
    {
        cin>>a>>b;
        p[i]={a*2,b*2};
    } 
    auto check=[&](int x0,int y0)->int
    {
        for(auto [x,y]:p)
        {
            if(x==x0||y==y0||y-y0==x-x0||y-y0==x0-x)
            {
                continue;
            }
            return 0;    //不满足    
        }
        return 1;        
    };
    auto [x0,y0]=p[0];
    for(int i=1;i<n;i++)
    {
        auto [x,y]=p[i];
        if(x0==x||y0==y||x0+y0==x+y||x0-x==y0-y)//在(x0,y0)点情况下所有点满足
        {
            continue;
        }
        if(check(x0+y0-y,y)||check(x0,y+(x-x0))||check(x0,y)||check(x0+y-y0,y)||check(x+y-y0,y0)||check(x,y0)||check(x-y+y0,y0)||check(x,y0-(x-x0))||check(x0,y-x+x0)||check(x,y0+(x-x0))||check((x0+x+y0-y)/2,(x0-x+y0+y)/2)||check((x0+x+y-y0)/2,(x-x0+y0+y)/2))
        {
            //如果这两个点不满足，判断中间能形成米子在他俩之间的点是否可以，总共12种情况
            cout<<"YES"<<"\n";
            return ;
        }else 
        {
            cout<<"NO"<<"\n";
            return ;
        }
    }
    cout<<"YES"<<'\n';

}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)
    {
        solve();
    }
    return 0;
}