Problem:
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The Maximum Number of Overlaps

Given a set of $N$ axis-aligned rectangular seals, find the number of overlapped seals on the region which has the maximum number of overlapped seals.

Input

The input is given in the following format.

$N$

$x1_1$ $y1_1$ $x2_1$ $y2_1$

$x1_2$ $y1_2$ $x2_2$ $y2_2$

:

$x1_N$ $y1_N$ $x2_N$ $y2_N$

($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left and the bottom-right corner of the $i$-th seal respectively.

Constraints

$ 1 \leq N \leq 100000 $

$ 0 \leq x1_i < x2_i \leq 1000 $

$ 0 \leq y1_i < y2_i \leq 1000 $

$ x1_i, y1_i, x2_i, y2_i$ are given in integers

Output

Print the maximum number of overlapped seals in a line.

Sample Input 1

2
0 0 3 2
2 1 4 3

Sample Output 1

2

Sample Input 2

2
0 0 2 2
2 0 4 2

Sample Output 2

1

Sample Input 3

3
0 0 2 2
0 0 2 2
0 0 2 2

Sample Output 3

3

Solution in C++14 (judge verdict: Accepted)

#include "bits/stdc++.h"

#define REP(i,num) for(int i=0;i<(num);++i)
#define ALL(c) c.begin(),c.end()
#define LOOP(i) while(i--)
#define PRINTALL(c) for(auto& x:c){cout<<x<<' ';}cout<<endl;
#define PAIRCOMP(c,comp) [](const pair<ll,ll>& lhs,const pair<ll,ll>& rhs){return lhs.c comp rhs.c;}

using namespace std;
using ll = long long;

constexpr ll atcoder_mod = 1e9+7;

template<typename T=int>
T in(){T x; cin >> x; return (x);}
template<typename T=int,typename C=vector<T>>
C vecin(int N){C x(N);REP(i,N){x[i]=in<T>();}return move(x);}

void vout(){cout << endl;}
template<typename Head,typename... Tail>
void vout(Head&& h,Tail&&... t){cout << ' ' << h;vout(forward<Tail>(t)...);}
void out(){cout << endl;}
template<typename Head,typename... Tail>
void out(Head&& h,Tail&&... t){cout << h;vout(forward<Tail>(t)...);}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N=in();
	vector<vector<pair<int,int>>> S(1001,vector<pair<int,int>>(1001,make_pair(0,0)));
	vector<vector<int>> A(1002,vector<int>(1002,0));
	LOOP(N){
		int LX=in(),LY=in(),RX=in(),RY=in();
		S[LY][LX].first++;
		S[LY][RX].second++;
		S[RY][LX].second++;
		S[RY][RX].first++;
	}

	int msum=0;
	REP(i,1001){
		REP(j,1001){
			A[i+1][j+1]+=A[i+1][j]+S[i][j].first-S[i][j].second;
			msum = max(msum,A[i+1][j+1]);
		}
	}

	REP(j,1002){
		REP(i,1001){
			A[i+1][j]+=A[i][j];
			msum = max(msum,A[i+1][j]);
		}
	}

	out(msum);
	return 0;
}