Problem:
Score : 800 points

Problem Statement

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

Partial Scores

500 points will be awarded for passing the test set satisfying 1 \leq D \leq 30.

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.

Sample Input 1

3 1
0 0
1 0
2 0

Sample Output 1

1

Sample Input 2

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

Sample Output 2

4

Sample Input 3

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

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
typedef long long ll;
typedef long double ld;
typedef complex<double> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

template<class T> void ckmin(T &a, T b) { a = min(a, b); }
template<class T> void ckmax(T &a, T b) { a = max(a, b); }

namespace io {
    // Source: StackOverflow
    
    template<class T> struct like_array : is_array<T>{};
    template<class T, size_t N> struct like_array<array<T,N>> : true_type{};
    template<class T> struct like_array<vector<T>> : true_type{};
    template<class T> bool is_like_array(const T& a) { return like_array<T>::value; }

    // I/O 
    
    void setIn(string s) { freopen(s.c_str(),"r",stdin); }
    void setOut(string s) { freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0);
        if (sz(s)) { setIn(s+".in"), setOut(s+".out"); }
    }
    
    // INPUT 
    
    // double input seems slow on CF
    void re(double& x) { string t; cin >> t; x = stod(t); } 
    void re(ld& x) { string t; cin >> t; x = stold(t); }
    template<class T> void re(T& x) { cin >> x; }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) { 
        re(first); re(rest...); 
    }
    template<class T1, class T2> istream& operator>>(istream& is, pair<T1,T2>& p) {
        is >> p.f >> p.s; return is;
    }
    template<class T> istream& operator>>(istream& is, vector<T>& a) {
        F0R(i,sz(a)) is >> a[i];
        return is;
    }
    
    // OUTPUT 
    
    template<class T> void pr(const T& x) { cout << x << '\n'; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) { 
        cout << first << ' '; pr(rest...); 
    }
    template<class T1, class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) {
        os << '{' << a.f << ", " << a.s << '}'; return os;
    }
    
    template<class T> ostream& printArray(ostream& os, const T& a, int SZ) {
        os << '{';
        F0R(i,SZ) {
            if (i) {
                os << ", ";
                if (is_like_array(a[i])) cout << "\n";
            }
            os << a[i];
        }
        os << '}';
        return os;
    }
    
    template<class T, size_t SZ> ostream& operator<<(ostream& os, const array<T,SZ>& a) {
        return printArray(os,a,SZ);
    }
    template<class T> ostream& operator<<(ostream& os, const vector<T>& a) {
        return printArray(os,a,sz(a));
    }
}

using namespace io;

int N,D;
int ans = MOD, cum[2][2001][2001];
int mx = 0;
vpi A,B;
map<pi,int> co;

int getSum(int k, int x0, int x1, int y0, int y1) {
    return cum[k][x1+1][y1+1]-cum[k][x0][y1+1]-cum[k][x1+1][y0]+cum[k][x0][y0];
}

bool ok(int x, int y, int mid) {
    if (getSum(1,x,x+mid,y,y+mid) != sz(B)) return 0;
    if (getSum(0,x+mid+1,x+D-1,y+mid+1,y+D-1) != 0) return 0;
    return 1;
}

void test(int x, int y) {
    int lo = 0, hi = D-1;
    while (lo < hi) {
        int mid = (lo+hi)/2;
        if (ok(x,y,mid)) hi = mid;
        else lo = mid+1;
    }
    ckmin(ans,lo);
}

int main() {
    // you should actually read the stuff at the bottom
    setIO(); re(N,D); 
    F0R(i,N) {
        int a,b; re(a,b);
        co[{a%D,b%D}] ++;
    }
    
    for (auto a: co) {
        int x = 1, y = 1;
        while (x*y < a.s) {
            if (x == y) y++;
            else x++;
        }
        if (mx < y) {
            A.clear(), B.clear();
            mx = y;
        }
        if (mx == y) {
            if (x == y) B.pb(a.f);
            else A.pb(a.f);
        }
    }
    for (auto t: A) {
        cum[0][t.f+1][t.s+1] ++;
        cum[0][t.f+D+1][t.s+1] ++;
        cum[0][t.f+1][t.s+1+D] ++;
        cum[0][t.f+1+D][t.s+1+D] ++;
    }
    for (auto t: B) {
        cum[1][t.f+1][t.s+1] ++;
        cum[1][t.f+D+1][t.s+1] ++;
        cum[1][t.f+1][t.s+1+D] ++;
        cum[1][t.f+1+D][t.s+1+D] ++;
    }
    F0R(k,2) FOR(i,1,2001) FOR(j,1,2001) cum[k][i][j] += cum[k][i-1][j]+cum[k][i][j-1]-cum[k][i-1][j-1];
    F0R(i,D) F0R(j,D) test(i,j);
    cout << ans+(mx-1)*D;
}

/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?), set tle
    * do smth instead of nothing and stay organized
*/