为了求矩形中的点的数量，首先可以想到的就是二维线段树或者二维树状数组，但是这道题的数据范围比较大，即使离散化之后还有200万多的点，会超空间。所以只能用一维的树状数组，可以先把所有点（包括查询矩形的四个顶点）按照第一关键字Y轴升序排序，第二关键字X轴升序排序。然后离散化，然后再扫一遍，如果是点就插入到树状数组中，如果是查询就求和，其值为右上角为改点，左下角为原点的矩形内的点，每个查询可以通过四个顶点的加减来考虑。有两点要考虑：1.要考虑边界，可以通过给点设值来排序。2.对N=0的情况要特判。
program p1935;
const
p:array[0..4] of longint=(4,1,3,5,2);
type
tree=record
x,y:longint;
end;
num=record
x,y,sx,pd,ls:longint;
end;
question=record
x1,x2,y1,y2:longint;
w:array[0..4] of longint;
end;
hzb=record
x,sx:longint;
end;
var
t:array[1..500000] of tree;
q:array[1..500000] of question;
x:array[0..3000000] of num;
h:array[0..3000000] of longint;
c:array[1..3000000] of longint;
len,n,m,i,j:longint;
procedure add(k1,k2,k3,k4:longint);
begin
inc(len);
x[len].x:=k1;
x[len].y:=k2;
x[len].sx:=k4;
x[len].pd:=k3;
end;
function bo(k1,k2:longint):boolean;
begin
if x[k1].y<x[k2].y then exit(true);
if x[k1].y>x[k2].y then exit(false);
if x[k1].pd<=1 then exit(true);
if x[k2].pd<=1 then exit(false);
if x[k1].x<x[k2].x then exit(true);
if x[k1].x>x[k2].x then exit(false);
if x[k1].pd<=x[k2].pd then exit(true);
exit(false);
end;
procedure quick(first,en:longint);
var
i,j:longint;
begin
{writeln(first,' ',en);}
i:=first; j:=en; x[0]:=x[first];
while i<j do begin
{writeln(x[i].x,' ',x[i].y,' ',x[i].pd);
writeln(x[j].x,' ',x[j].y,' ',x[j].pd);
writeln(x[0].x,' ',x[0].y,' ',x[0].pd);}
while (i<j) and (bo(0,j)=true) do dec(j);
if i<j then x[i]:=x[j];
while (i<j) and (bo(i,0)=true) do inc(i);
if i<j then x[j]:=x[i];
end;
x[i]:=x[0];
if i>first+1 then quick(first,i-1);
if i<en-1 then quick(i+1,en);
end;
procedure sort(first,en:longint);
var
i,j:longint;
begin
i:=first; j:=en; h[0]:=h[i];
while i<j do begin
while (i<j) and ((x[h[j]].x>x[h[0]].x) or ((x[h[j]].x=x[h[0]].x)) and (p[x[h[j]].pd]>=p[x[h[0]].pd])) do dec(j);
if i<j then h[i]:=h[j];
while (i<j) and ((x[h[i]].x<x[h[0]].x) or ((x[h[i]].x=x[h[0]].x)) and (p[x[h[i]].pd]<p[x[h[0]].pd])) do inc(i);
if i<j then h[j]:=h[i];
end;
h[i]:=h[0];
if i>first+1 then sort(first,i-1);
if i<en-1 then sort(i+1,en);
end;
function lowbit(k:longint):longint;
begin
exit(k and -k);
end;
procedure add(k:longint);
var
i,j:longint;
begin
i:=k;
while i<=len do begin
inc(c[i]);
i:=i+lowbit(i);
end;
end;
function total(k:longint):longint;
var
i,j,num:longint;
begin
i:=k; num:=0;
while i>0 do begin
num:=num+c[i];
i:=i-lowbit(i);
end;
exit(num);
end;
begin
readln(n,m); len:=0;
fillchar(x,sizeof(x),0);
for i:=1 to n do begin
readln(t[i].x,t[i].y);
add(t[i].x,t[i].y,2,i);
end;
if n=0 then begin
for i:=1 to m do
writeln(0);
exit;
end;
for i:=1 to m do begin
readln(q[i].x1,q[i].y1,q[i].x2,q[i].y2);
add(q[i].x1,q[i].y1,1,i);
add(q[i].x2,q[i].y2,3,i);
add(q[i].x1,q[i].y2,4,i);
add(q[i].x2,q[i].y1,0,i);
end;
quick(1,len);
fillchar(c,sizeof(c),0);
for i:=1 to len do
h[i]:=i;
sort(1,len);
for i:=1 to len do
x[h[i]].ls:=i;
for i:=1 to len do
if x[i].pd=2 then
add(x[i].ls)
else
q[x[i].sx].w[x[i].pd]:=total(x[i].ls);
for i:=1 to m do begin
writeln(q[i].w[3]+q[i].w[1]-q[i].w[0]-q[i].w[4]);
end;
end.
