Captain Marmot (Codeforces Round #271 div2) C

Captain MarmotTime limit:MS Memory Limit:262144KB 64bit IO Format:%i64d &%i64u SubmitStatus

Description

Captain Marmot wants to prepare a huge and important battle against his enemy, Captain Snake. For this battle he had N Regiments, each consisting of 4 moles.

Initially, each mole i (1?≤? I? ≤?4N) is placed at some position (x_{i,? Yi)} in the Cartesian plane. Captain Marmot wants to move some moles to make the regiments compact, if it ' s possible.

Each mole  i  has a home placed at the Position  ( a _{i ,? b i )} . Moving this mole one time means rotating his position point  ( x _{i ,? Y i )} 90  degrees counter-clockwise around it's home point  ( a _{i ,? b i )} .

A regiment is compact only if the position points of the 4 moles form a square with Non-zero area.

Help Captain Marmot to find out for each regiment the minimal number of moves required to make that regiment compact, if I T ' s possible.

Input

The first line contains one integer n (1?≤? N? ≤?100), the number of regiments.

The Next 4 n  lines Contain 4  integers  x _i ,   y _i ,   a _i ,   b _i   (?-? 10^{4?≤? X _{i ,? Y i ,? A i ,? b i ? ≤?104}} ).

Output

Print n lines to the standard output. If the Regiment I can be made compact, the I--th line should contain one integer, the minimal number of R Equired moves. Otherwise, on the i-th line Print "-1" (without quotes).

Sample Input

Input

41 1 0 0-1 1 0 0-1 1 0 01-1 0 01 1 0 0-2 1 0 0-1 1 0 01-1 0 01 1 0 0-1 1 0 0-1 1 0 0-1 1 0 02 2 0 1-1 0 0-23 0 0-2-1 1 -2 0

Output

1-133

Hint

In the first regiment we can move once the second or the third mole.

We can ' t make the Second Regiment compact.

In the third regiment, from the last 3 moles We can move once one and twice another one.

In the Fourth regiment, we can move twice the first mole and once the third mole.

#include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include < iostream> #include <sstream> #include <algorithm> #include <set> #include <queue> #include <stack> #include <map>using namespace std;typedef long long ll;const int inf=0x3f3f3f3f;const double pi= ACOs ( -1.0); #define Lson L,mid,rt<<1#define Rson mid+1,r,rt<<1|1struct Node {int x, y;} rec[10][10],cen[10];    ll dis (struct node a,struct Node B) {ll xx,yy;    LL Res;    xx= (a.x-b.x) * (a.x-b.x);    yy= (A.Y-B.Y) * (A.Y-B.Y);    Res=xx+yy; return res;}    LL bian[6];//ll xie[2];void judge () {int i,j,k,l;    int ans=inf;                     For (i=0, i<4; i++) for (j=0; j<4; j + +) for (k=0; k<4; k++) for (l=0; l<4; l++) {                    Bian[0]=dis (rec[i][0],rec[j][1]);                    Bian[1]=dis (rec[j][1],rec[k][2]);                    Bian[2]=dis (Rec[k][2],rec[l][3]); Bian[3]=dis (Rec[l][3],Rec[i][0]);                    Bian[4]=dis (rec[i][0],rec[k][2]);                    Bian[5]=dis (Rec[j][1],rec[l][3]);                    for (int ii=0;ii<6;ii++)//printf ("bian[%d]==%d", Ii,bian[ii]);                    Sort (bian,bian+6); if (bian[0]==0| | bian[1]==0| | bian[2]==0| | bian[3]==0| | bian[4]==0| |                    bian[5]==0) continue; else if (bian[0]==bian[1]&&bian[1]==bian[2]&&bian[2]==bian[3]&&bian[3]==bian[0]&                    &AMP;XIE[0]==XIE[1]&AMP;&AMP;BIAN[0]*2==XIE[0])//Ans=min (ANS,I+J+K+L); else if (bian[0]==bian[1]&&bian[1]==bian[2]&&bian[2]==bian[3]&&bian[3]==bian[0]&                &AMP;BIAN[4]==BIAN[5]&AMP;&AMP;BIAN[0]*2==BIAN[4]) ans=min (ans,i+j+k+l);    } if (Ans!=inf) printf ("%d\n", ans); else printf (" -1\n");}    int main () {int t,i,j;    scanf ("%d", &t); while (t--) {for (i=0; i<4;            i++) {scanf ("%d%d", &rec[0][i].x,&rec[0][i].y);            scanf ("%d%d", &cen[i].x,&cen[i].y);            rec[1][i].x=cen[i].x-(REC[0][I].Y-CEN[I].Y);            rec[1][i].y=cen[i].y+ (rec[0][i].x-cen[i].x);            rec[2][i].x=cen[i].x-(rec[0][i].x-cen[i].x);            rec[2][i].y=cen[i].y-(REC[0][I].Y-CEN[I].Y);            rec[3][i].x=cen[i].x+ (REC[0][I].Y-CEN[I].Y);        rec[3][i].y=cen[i].y-(rec[0][i].x-cen[i].x);    } judge (); } return 0;}

Captain Marmot (Codeforces Round #271 div2) C