#include <set>

#include <map>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <string>

#include <vector>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define debug puts("here")

#define rep(i,n) for(int i=0;i<n;i++)

#define rep1(i,n) for(int i=1;i<=n;i++)

#define REP(i,a,b) for(int i=a;i<=b;i++)

#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)

#define pb push_back

#define RD(n) scanf("%d",&n)

#define RD2(x,y) scanf("%d%d",&x,&y)

#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)

#define All(vec) vec.begin(),vec.end()

#define MP make_pair

#define PII pair<int,int>

#define PQ priority_queue

#define cmax(x,y) x = max(x,y)

#define cmin(x,y) x = min(x,y)

#define Clear(x) memset(x,0,sizeof(x))

/*

#pragma comment(linker, "/STACK:1024000000,1024000000")

int size = 256 << 20; // 256MB

char *p = (char*)malloc(size) + size;

__asm__("movl %0, %%esp\n" :: "r"(p) );

*/

char IN;

bool NEG;

inline void Int(int &x){

    NEG = 0;

    while(!isdigit(IN=getchar()))

        if(IN=='-')NEG = 1;

    x = IN-'0';

    while(isdigit(IN=getchar()))

        x = x*10+IN-'0';

    if(NEG)x = -x;

}

inline void LL(ll &x){

    NEG = 0;

    while(!isdigit(IN=getchar()))

        if(IN=='-')NEG = 1;

    x = IN-'0';

    while(isdigit(IN=getchar()))

        x = x*10+IN-'0';

    if(NEG)x = -x;

}

/******** program ********************/

PII p[22];

int px[22],py[22];

int a[100],n;

int dp[2][1<<21];

int main(){

#ifndef ONLINE_JUDGE

    freopen("sum.in","r",stdin);

    //freopen("sum.out","w",stdout);

#endif

    while(RD(n),~n){

        rep1(i,n)

            RD2(p[i].first,p[i].second);

        sort(p+1,p+n+1);

        rep1(i,n)

            px[i] = p[i].first , py[i] = p[i].second;

        int tp = 0;

        rep1(i,n) // 处理出所有可以组成的正方形

        REP(j,i+1,n)if(px[i]==px[j])

        REP(k,j+1,n)if(py[i]==py[k]||py[j]==py[k])

        REP(l,k+1,n)if(px[k]==px[l])

        if( (py[i]==py[k]&&py[j]==py[l])||(py[i]==py[l]&&py[j]==py[k]) )

        if( abs(px[i]-px[k])==abs(py[i]-py[j]) )

            a[++tp] = (1<<(i-1))|(1<<(j-1))|(1<<(k-1))|(1<<(l-1));

        Clear(dp);

        int all = 1<<n;

        int ans = 0;

        rep1(i,tp) // 状态压缩

            for(int sta=all-1;sta>=0;sta--){

                dp[i&1][sta] = dp[!(i&1)][sta];

                if( (a[i]&sta)!=a[i] )continue;

                cmax( dp[i&1][sta] , dp[!(i&1)][sta-a[i]]+4 );

                cmax(ans,dp[i&1][sta]);

            }

        cout<<ans<<endl;

    }

    return 0;

}