Problem: Niwango-kun, an employee of Dwango Co., Ltd., likes Niconico TV-chan, so he collected a lot of soft toys of her and spread them on the floor.

Niwango-kun has N black rare soft toys of Niconico TV-chan and they are spread together with ordinary ones. He wanted these black rare soft toys to be close together, so he decided to rearrange them.

In an infinitely large two-dimensional plane, every lattice point has a soft toy on it. The coordinates (x_i,y_i) of N black rare soft toys are given. All soft toys are considered to be points (without a length, area, or volume).

He may perform the following operation arbitrarily many times:

* Put an axis-aligned square with side length D, rotate the square by 90 degrees with four soft toys on the four corners of the square. More specifically, if the left bottom corner's coordinate is (x, y), rotate four points (x,y) \rightarrow (x+D,y) \rightarrow (x+D,y+D) \rightarrow (x,y+D) \rightarrow (x,y) in this order. Each of the four corners of the square must be on a lattice point.



Let's define the scatteredness of an arrangement by the minimum side length of an axis-aligned square enclosing all black rare soft toys. Black rare soft toys on the edges or the vertices of a square are considered to be enclosed by the square.

Find the minimum scatteredness after he performs arbitrarily many operations.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq D \leq 1000
* 0 \leq x_i, y_i \leq 10^9
* Given coordinates are pairwise distinct
* All numbers given in input are integers

Input

Input is given from Standard Input in the following format:


N D
x_1 y_1
:
x_N y_N


Output

Print the answer.

Examples

Input

3 1
0 0
1 0
2 0


Output

1


Input

19 2
1 3
2 3
0 1
1 1
2 1
3 1
4 4
5 4
6 4
7 4
8 4
8 3
8 2
8 1
8 0
7 0
6 0
5 0
4 0


Output

4


Input

8 3
0 0
0 3
3 0
3 3
2 2
2 5
5 2
5 5


Output

4

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase

/*
cout<<fixed<<setprecision(20);
cin.tie(0);
ios::sync_with_stdio(false);
*/
const llint mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	llint n,D,i,j,x,y;cin>>n>>D;
	static int kaz[1001][1001]={0};
	for(i=0;i<n;i++){cin>>x>>y;kaz[y%D][x%D]++;}
	int syo=0;
	for(y=0;y<D;y++){
		for(x=0;x<D;x++){maxeq(syo,kaz[y][x]);}
	}
	syo=(int)(sqrt(syo)+0.999999);
	//syoが最大のあれ トーラスに気を付ける
	//kazの意味を変える
	for(y=0;y<D;y++){
		for(x=0;x<D;x++){
			//cerr<<kaz[y][x]<<" ";
			if(kaz[y][x]>syo*(syo-1)){kaz[y][x]=2;}
			else if(kaz[y][x]>(syo-1)*(syo-1)){kaz[y][x]=1;}
			else {kaz[y][x]=0;}
		}//cerr<<endl;
	}
	//区間内に2や1があるかどうかを累積和で調べる
	static int niwa[2001][2001]={0};
	static int itwa[2001][2001]={0};
	for(y=0;y<D+D;y++){
		for(x=0;x<D+D;x++){
			niwa[y+1][x+1]=niwa[y+1][x];
			itwa[y+1][x+1]=itwa[y+1][x];
			if(kaz[y%D][x%D]==2){niwa[y+1][x+1]++;}
			if(kaz[y%D][x%D]>=1){itwa[y+1][x+1]++;}
		}
	}
	for(y=0;y<D+D;y++){
		for(x=0;x<D+D;x++){
			niwa[y+1][x+1]+=niwa[y][x+1];
			itwa[y+1][x+1]+=itwa[y][x+1];
		}
	}
	int bmin=0,bmax=D;
	int nisu=niwa[D][D];//全体の2の数
	while(bmax-bmin>1){
		int b=(bmin+bmax)/2;
		bool can=0;
		for(y=0;y<D;y++){
			for(x=0;x<D;x++){
				if(nisu==niwa[y+b][x+b]+niwa[y][x]-niwa[y+b][x]-niwa[y][x+b]&&
				0==itwa[y+D][x+D]+itwa[y+b][x+b]-itwa[y+D][x+b]-itwa[y+b][x+D])
				{can=1;break;}
			}
		}
		if(can){bmax=b;}
		else{bmin=b;}
	}
	cout<<(syo-1)*D+bmax-1<<endl;
	return 0;
}