#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+10, inf=1e9;
LL n, ans, sum, x[N], y[N], m[N], v[N], st[N];
void solve(LL lx, LL rx, LL ly, LL ry) {
    sum=0;
    for(int i=1;i<=n;i++) {
        if(lx <= x[i] && rx >= x[i] && ly <= y[i] && ry >= y[i]) {v[i]=0; continue;} //如果不在范围内就不管 
        if(lx <= y[i] && rx >= y[i] && ly <= x[i] && ry >= x[i]) {v[i]=1; sum+=m[i];} //如果在则加上 
        else return ; //不可能更优，排除 
    }
    if(sum < ans) {ans=sum; for(int i=1;i<=n;i++) st[i]=v[i];} //更新答案 
}
int main() {
    scanf("%d", &n);
    LL lx=inf, rx=0, ly=inf, ry=0;
    //为了使lx!=ly,rx!=ry,我们定义min(x[i],y[i])为横坐标，max(x[i], y[i])为纵坐标 
    //即lx<ly, rx<ry 
    //lx表示最小的横坐标，ly表示最大的纵坐标中最小的 (rx,ry同理)
    //即求出边界的4个点
    //从右至左，上至下依次为
    //(lx, rx), (lx, ry)
    //(ly, rx), (ly, ry) 
    for(int i=1;i<=n;i++) {
        scanf("%d%d%d", &x[i], &y[i], &m[i]);
        lx=min(lx, min(x[i], y[i])), rx=max(rx, min(x[i], y[i]));
        ly=min(ly, max(x[i], y[i])), ry=max(ry, max(x[i], y[i]));
    }
    printf("%lld ", 2*(rx+ry-lx-ly));
    ans=inf;
    //尝试是否可以翻折  
    //对角线逐一对应 （考虑将所有点翻折到y=x的一边） 
    solve(lx, rx, ly, ry); 
    solve(lx, ry, ly, rx);
    solve(ly, rx, lx, ry); 
    solve(ly, ry, lx, rx);
    printf("%lld\n", ans);
    for(int i=1;i<=n;i++) printf("%lld", st[i]);
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+10, inf=1e9;
LL n, ans, sum, x[N], y[N], m[N], v[N], st[N];
void solve(LL lx, LL rx, LL ly, LL ry) {
	sum=0;
	for(int i=1;i<=n;i++) {
		if(lx<=x[i]&&rx>=x[i]&&ly<=y[i]&&ry>=y[i]) {v[i]=0; continue;} //如果不在范围内就不管 
		if(lx<=y[i]&&rx>=y[i]&&ly<=x[i]&&ry>=x[i]) {v[i]=1; sum+=m[i];} //如果在则加上 
		else return ; //不可能更优，排除 
	}
	if(sum<ans) {ans=sum; for(int i=1;i<=n;i++) st[i]=v[i];} //更新答案 
}
int main() {
	scanf("%d", &n);
	LL lx=inf, rx=0, ly=inf, ry=0;
	//为了使lx!=ly,rx!=ry,我们定义min(x[i],y[i])为横坐标，max(x[i], y[i])为纵坐标 
	//即lx<ly, rx<ry 
	//lx表示最小的横坐标，ly表示最大的纵坐标中最小的 (rx,ry同理)
	//即求出边界的4个点
	//从右至左，上至下依次为
	//(lx, rx), (lx, ry)
	//(ly, rx), (ly, ry) 
	for(int i=1;i<=n;i++) {
		scanf("%d%d%d", &x[i], &y[i], &m[i]);
		lx=min(lx, min(x[i], y[i])), rx=max(rx, min(x[i], y[i]));
		ly=min(ly, max(x[i], y[i])), ry=max(ry, max(x[i], y[i]));
	}
	printf("%lld ", 2*(rx+ry-lx-ly));
	ans=inf;
	//尝试是否可以翻折  
	//对角线逐一对应 （考虑将所有点翻折到y=x的一边） 
	solve(lx, rx, ly, ry); 
	solve(lx, ry, ly, rx);
	solve(ly, rx, lx, ry); 
	solve(ly, ry, lx, rx);
	printf("%lld\n", ans);
	for(int i=1;i<=n;i++) printf("%lld", st[i]);
	return 0;
}