/*

 凸包（稳定凸包）

 题意：给出一些点，这些点要么是凸包的顶点要么是边上的。

 证明每条边上都至少有3个点。

 */

 #include<stdio.h>

 #include<string.h>

 #include<stdlib.h>

 #include<algorithm>

 #include<iostream>

 #include<queue>

 #include<map>

 #include<stack>

 #include<set>

 #include<math.h>

 using namespace std;

 typedef long long int64;

 //typedef __int64 int64;

 typedef pair<int64,int64> PII;

 #define MP(a,b) make_pair((a),(b))

 const int maxn = ;

 const int inf = 0x7fffffff;

 const double pi=acos(-1.0);

 const double eps = 1e-;

 struct Point {

     double x,y;

     bool operator < ( const Point p ) const {

         return y<p.y||(y==p.y&&x<p.x);

     }

 }res[ maxn ],pnt[ maxn ];

 bool flag[ maxn ][ maxn ];//f[i][j]:点ij之间是否还有点

 bool ok[ maxn ];//ok[i]:i是否是凸包的顶点

 double xmult( Point sp,Point ep,Point op ){

     return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);

 }

 double dis2( Point a,Point b ){

     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);

 }

 double dis( Point a,Point b ){

     return sqrt( dis2(a,b) );

 } 

 bool PointOnLine( Point a,Point L,Point R ){

     return ( fabs(xmult( a,L,R ))<eps )&&( (a.x-L.x)*(a.x-R.x)<eps )&&( (a.y-L.y)*(a.y-R.y)<eps );

 }

 int Garham( int n ){

     int top = ;

     sort( pnt,pnt+n );

     if( n== ) return ;

     else res[  ] = pnt[  ];

     if( n== ) return ;

     else res[  ] = pnt[  ];

     if( n== ) return ;

     else res[  ] = pnt[  ];

     for( int i=;i<n;i++ ){

         while( top>&&xmult( res[top],res[top-],pnt[i] )>= )

             top--;

         res[ ++top ] = pnt[ i ];

     }

     int tmp = top;

     res[ ++top ] = pnt[ n- ];

     for( int i=n-;i>=;i-- ){

         while( top!=tmp&&xmult( res[top],res[top-],pnt[i] )>= )

             top--;

         res[ ++top ] = pnt[ i ];

     }

     return top;

 }

 int main(){

     int T;

     scanf("%d",&T);

     while( T-- ){

         int n;

         scanf("%d",&n);

         for( int i=;i<n;i++ )

             scanf("%lf%lf",&pnt[i].x,&pnt[i].y);

         if( n<= ){

             puts("NO");

             continue;

         }//至少要6个点

         int cnt = Garham( n );

         memset( ok,false,sizeof( ok ) );

         memset( flag,false,sizeof( flag ) );

         for( int i=;i<n;i++ ){

             for( int j=;j<cnt;j++ ){

                 if( pnt[i].x==res[j].x&&pnt[i].y==res[j].y ){

                     ok[ i ] = true;

                     break;

                 }

             }

         }

         for( int i=;i<n;i++ ){

             if( !ok[i] ){

                 for( int j=;j<cnt;j++ ){

                     if( PointOnLine( pnt[i],res[j],res[(j+)%cnt]) ){

                         flag[ j ][ (j+)%cnt ] = true;

                     }

                 }

             }

         }

         bool ans = true;

         for( int i=;i<cnt;i++ ){

             if( flag[ i ][ (i+)%cnt ]==false ){

                 ans = false;

                 break;

             }

         }

         if( ans ) printf("YES\n");

         else puts("NO");

     }

     return ;

 }