#include<bits/stdc++.h>
#define int long long
using namespace std;
void ts(){cout<<"IAKIOI\n";}
inline int read(){
	int n=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		n=n*10+ch-'0';
		ch=getchar();
	}
	return n*f;
}
bool Mbe;
int ty[1000005],d[1000005],a[1000005],b[1000005];
int lsh[1000005];
int l;
multiset<int> ga[1000005],gb[1000005],ha[1000005],hb[1000005];
struct node{
	int val=(1ll<<60);
	int sa=(1ll<<60),sb=(1ll<<60),ta=(1ll<<60),tb=(1ll<<60);
	//x 来自左边, y 来自右边.
	//答案就是 bx+by.
	//否则答案就是 ax+ay.
	//维护区间内每个集合最小的 a 和 b.
	//这个区间的 max(ax+ay,bx+by)
	//ax-bx=by-ay
	
}seg[1000005*4];
void upd(int x){
	seg[x].val=min(seg[x*2].val,seg[x*2+1].val);
	seg[x].val=min(seg[x].val,seg[x*2].sb+seg[x*2+1].tb);
	seg[x].val=min(seg[x].val,seg[x*2].ta+seg[x*2+1].sa);
	seg[x].sa=min(seg[x*2].sa,seg[x*2+1].sa);
	seg[x].sb=min(seg[x*2].sb,seg[x*2+1].sb);
	seg[x].ta=min(seg[x*2].ta,seg[x*2+1].ta);
	seg[x].tb=min(seg[x*2].tb,seg[x*2+1].tb);
}
void modify(int id,int l,int r,int x){
	if(l==r){
		seg[id].sa=(ga[x].empty()?(1ll<<60):*ga[x].begin());
		seg[id].sb=(gb[x].empty()?(1ll<<60):*gb[x].begin());
		seg[id].ta=(ha[x].empty()?(1ll<<60):*ha[x].begin());
		seg[id].tb=(hb[x].empty()?(1ll<<60):*hb[x].begin());
		if(seg[id].sa+seg[id].ta<2000000002)seg[id].val=seg[id].sa+seg[id].ta;
		else seg[id].val=(1ll<<60);
		//cout<<l<<" "<<r<<" "<<x<<" "<<seg[id].sa<<" "<<seg[id].sb<<" "<<seg[id].ta<<" "<<seg[id].tb<<"\n";
		return;
	}
	int mid=(l+r)/2;
	if(x<=mid)modify(id*2,l,mid,x);
	else modify(id*2+1,mid+1,r,x);
	upd(id);
}
bool Med;
signed main(){
	//cout<<(&Mbe-&Med)/1048576.00<<"\n";
	int t=read();
	for(int i=1;i<=t;i++){
		ty[i]=read(),d[i]=read(),a[i]=read(),b[i]=read();
		if(!d[i])lsh[++l]=a[i]-b[i];
		else lsh[++l]=b[i]-a[i];
	}
	sort(lsh+1,lsh+l+1);
	int m=unique(lsh+1,lsh+l+1)-lsh-1;
	for(int i=1;i<=t;i++){
		if(ty[i]==1){
			//加入
			if(!d[i]){
				//加入第 1 个集合
				int x=lower_bound(lsh+1,lsh+m+1,a[i]-b[i])-lsh;
				ga[x].insert(a[i]),gb[x].insert(b[i]);
				modify(1,1,m,x);
			}
			else{
				int x=lower_bound(lsh+1,lsh+m+1,b[i]-a[i])-lsh;
				ha[x].insert(a[i]),hb[x].insert(b[i]);
				modify(1,1,m,x);
			}
		}
		else{
			if(!d[i]){
				int x=lower_bound(lsh+1,lsh+m+1,a[i]-b[i])-lsh;
				ga[x].erase(ga[x].find(a[i]));
				gb[x].erase(gb[x].find(b[i]));
				modify(1,1,m,x);
			}
			else{
				int x=lower_bound(lsh+1,lsh+m+1,b[i]-a[i])-lsh;
				ha[x].erase(ha[x].find(a[i]));
				hb[x].erase(hb[x].find(b[i]));
				modify(1,1,m,x);
			}
		}
		if(seg[1].val>2000000002){printf("-1\n");}
		else printf("%lld\n",seg[1].val);
	}
	//ax+ay>=bx+by
	//ax-bx>=by-ay
	//维护 a-b 和 b-a.
	//然后考虑 [l,r] 里要找出 max(ax+ay,bx+by)
	//
	return 0;
}
//look at my code
//my code is amazing