//BZOJ4066 "简单"题

 //维护N*N矩形，初始全为0，支持两种操作：

 //1.将格子(x,y)的数字加上A

 //2.求(x1,y1)到(x2,y2)这个矩形内的数字和

 //3.结束程序

 //由于平衡性，每5000次插入就暴力重构一次

 #include<algorithm>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<cmath>

 using namespace std;

 int D;

 struct kdnode{

     int ls,rs,num,v,d[],mi[],mx[];

     int &operator [](int x){

         return d[x];

     }

     friend bool operator <(kdnode a,kdnode b){

         return a[D]<b[D];

     }

 }t[],rb[],s;

 int n,op,x1,yy,x2,y2,v,rt=,tot=,R=,ans=;

 bool inside(int x1,int yy,int x2,int y2,int x3,int y3,int x4,int y4){

     return x1<=x3&&x2>=x4&&yy<=y3&&y2>=y4;

 }

 bool outside(int x1,int yy,int x2,int y2,int x3,int y3,int x4,int y4){

     return x1>x4||x2<x3||yy>y4||y2<y3;

 }

 void pushup(int u){

     int l=t[u].ls,r=t[u].rs;

     for(int i=;i<=;i++){

         t[u].mi[i]=t[u].mx[i]=t[u][i];

         if(l)t[u].mi[i]=min(t[u].mi[i],t[l].mi[i]);

         if(l)t[u].mx[i]=max(t[u].mx[i],t[l].mx[i]);

         if(r)t[u].mi[i]=min(t[u].mi[i],t[r].mi[i]);

         if(r)t[u].mx[i]=max(t[u].mx[i],t[r].mx[i]);

     }

     t[u].num=t[l].num+t[r].num+t[u].v;

 }

 int query(int u,int x1,int yy,int x2,int y2){

     int ret=;

     if(!u)return ;

     if(inside(x1,yy,x2,y2,t[u].mi[],t[u].mi[],t[u].mx[],t[u].mx[]))return t[u].num;

     if(outside(x1,yy,x2,y2,t[u].mi[],t[u].mi[],t[u].mx[],t[u].mx[]))return ;

     if(inside(x1,yy,x2,y2,t[u][],t[u][],t[u][],t[u][]))ret+=t[u].v;

     ret+=query(t[u].ls,x1,yy,x2,y2)+query(t[u].rs,x1,yy,x2,y2);

     return ret;

 }

 void ins(int &u,bool d){

     if(!u){

         u=++tot;

         t[u][]=t[u].mi[]=t[u].mx[]=s[];

         t[u][]=t[u].mi[]=t[u].mx[]=s[];

     }

     if(s[]==t[u][]&&s[]==t[u][]){

         t[u].v+=s.v;

         t[u].num+=s.v;

         return;

     }

     if(s[d]<t[u][d])ins(t[u].ls,d^);

     else ins(t[u].rs,d^);

     pushup(u);

 }

 int rebuild(int l,int r,bool d){

     if(l>r)return ;

     int mid=(l+r)/;

     D=d;

     nth_element(rb+l,rb+mid,rb+r+);

     t[mid]=rb[mid];

     t[mid].ls=rebuild(l,mid-,d^);

     t[mid].rs=rebuild(mid+,r,d^);

     pushup(mid);

     return mid;

 }

 int main(){

     scanf("%d",&n);

     for(;;){

         scanf("%d",&op);

         if(op==){

             scanf("%d%d%d",&x1,&yy,&v);

             x1^=ans;

             yy^=ans;

             v^=ans;

             s[]=x1;

             s[]=yy;

             s.num=s.v=v;

             ins(rt,);

             if(tot==R){

                 for(int j=;j<=tot;j++){

                     rb[j]=t[j];

                 }

                 rt=rebuild(,tot,);

                 R+=;

             }

         }else if(op==){

             scanf("%d%d%d%d",&x1,&yy,&x2,&y2);

             x1^=ans;

             yy^=ans;

             x2^=ans;

             y2^=ans;

             printf("%d\n",(ans=query(rt,x1,yy,x2,y2)));

         }else break;

     }

     return ;

 }