**链接：****传送门 **
题意：略
思路：数据量很小，直接暴力所有线段
/*************************************************************************
> File Name: hdu2150.cpp
> Author: WArobot
> Blog: http://www.cnblogs.com/WArobot/
> Created Time: 2017年05月07日 星期日 23时09分58秒
************************************************************************/
#include<bits/stdc++.h>
using namespace std;
#define eps 1e-10
struct point{ double x,y; };
bool inter(point a,point b,point c,point d){
if( min(a.x,b.x) > max(c.x,d.x) ||
min(a.y,b.y) > max(c.y,d.y) ||
min(c.x,d.x) > max(a.x,b.x) ||
min(c.y,d.y) > max(a.y,b.y) ) return 0;
double h , i , j , k;
h = (b.x - a.x)*(c.y - a.y) - (b.y - a.y)*(c.x - a.x);
i = (b.x - a.x)*(d.y - a.y) - (b.y - a.y)*(d.x - a.x);
j = (d.x - c.x)*(a.y - c.y) - (d.y - c.y)*(a.x - c.x);
k = (d.x - c.x)*(b.y - c.y) - (d.y - c.y)*(b.x - c.x);
return h*i<=eps && j*k<=eps;
}
int main(){
int N,num[31];
point pi[31][101];
while(~scanf("%d",&N)){
for(int i=0;i<N;i++){
scanf("%d",&num[i]);
for(int j=0;j<num[i];j++){
scanf("%lf%lf",&pi[i][j].x,&pi[i][j].y);
}
}
bool ok = false;
for(int i=0;i<N;i++){
for(int j=i+1;j<N;j++){
for(int k=0;k<num[i]-1;k++){
for(int s=0;s<num[j]-1;s++){
if( inter(pi[i][k],pi[i][k+1],pi[j][s],pi[j][s+1])){
ok = true; break;
}
}
}
}
}
if(ok) printf("Yes\n");
else printf("No\n");
}
return 0;
}