/**************************************************************
    Problem: 1100
    User: zrts
    Language: C++
    Result: Accepted
    Time:2064 ms
    Memory:9456 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<cmath>
//  by zrt
//  problem:
//  无论你在什么时候开始，重要的是开始以后就不要停止。
using namespace std ;
typedef long long ll ;
const double eps(1e-5) ;
const int inf(0x7fffffff) ;
#define MAXN 100500
int tt;
int n;
int x[MAXN],y[MAXN];
int r[4*MAXN];
struct N{
    bool type;
    double a;
    int c;
    friend bool operator == ( N a,N b){
        return a.type==b.type&&fabs(a.a-b.a)<eps&&a.c+b.c!=0;
    }
};
N a[4*MAXN];
int main(){
    #ifdef LOCAL
    freopen("in.txt","r",stdin) ;
    freopen("out.txt","w",stdout) ;
    #endif
    scanf("%d",&tt);
    while(tt--){
        scanf("%d",&n);
        int ans(0);
        for(int i=0;i<n;i++){
            scanf("%d%d",&x[i],&y[i]);
        }
        for(int i=0;i<n;i++){// bian
            a[i<<1|1].type=1;
            a[i<<1|1].c=1;
            a[i<<1|1].a=(x[i]-x[(i+1)%n])*1LL*(x[i]-x[(i+1)%n])+(y[i]-y[(i+1)%n])*1LL*(y[i]-y[(i+1)%n]);
        }
        for(int i=0,x1,y1,x2,y2;i<n;i++){
            a[i<<1].type=0;
            x1=x[(i+1)%n]-x[i];
            y1=y[(i+1)%n]-y[i];
            x2=x[((i-1+n)%n)]-x[i];
            y2=y[((i-1+n)%n)]-y[i];
            long long sinp=x1*1LL*y2-x2*1LL*y1;
            if(sinp+eps>0) a[i<<1].c=1;
            else a[i<<1].c=-1; 
            a[i<<1].a=fabs(sinp)/sqrt(x1*1LL*x1+y1*1LL*y1)/sqrt(x2*1LL*x2+y2*1LL*y2);
        }
        int lim=2*n;
        for(int i=0;i<lim;i++){
            a[i+lim]=a[i];
        }
        lim<<=1;
        int mx(-1),id(-1);
        for(int i=0;i<lim;i++){
        //  printf("#%d %d %.3f\n",i,a[i].type,a[i].a);
            r[i]=0;
            if(i<mx){
                r[i]=min(mx-i,r[id-(i-id)]);
            }
            while(i+r[i]<lim&&i-r[i]>=0&&a[i+r[i]]==a[i-r[i]]) r[i]++;
            if(i+r[i]>mx){
                mx=i+r[i];
                id=i;
            }
            if((r[i]<<1)-1>n<<1){
                ans++;
            }
        }
        printf("%d\n",ans>>1);
    }
    return 0 ;
}