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 <bits/stdc++.h>
#define ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 300010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

int N, D;
int S[1010][1010];
int E[2010][2010];
int F[2010][2010];

bool ok(ll m) {
	ll sv = m / D, lv = m / D + 1;
	ll l = 0;
	rep(i, 0, D) if(i + D * sv < m) l++;
	memset(E, 0, sizeof(E)); memset(F, 0, sizeof(F));
	int lcnt = 0;
	rep(i, 0, D) {
		rep(j, 0, D) {
			if(S[i][j] > lv * lv) return false;
		}
	}
	rep(i, 0, D * 2) {
		rep(j, 0, D * 2) {
			if(S[i % D][j % D] > lv * sv) {
				lcnt++;
				E[i + 1][j + 1]++;
			}
			if(S[i % D][j % D] > sv * sv) {
				F[i + 1][j + 1]++;
			}
			E[i + 1][j + 1] += E[i][j + 1] + E[i + 1][j] - E[i][j];
			F[i + 1][j + 1] += F[i][j + 1] + F[i + 1][j] - F[i][j];
		}
	}
	lcnt /= 4;
	// debug(sv, lv, l);
	// rep(i, 0, D) debug(vi(S[i], S[i] + D));
	// rep(i, 0, 2 * D + 1) debug(vi(E[i], E[i] + 2 * D + 1));
	// rep(i, 0, 2 * D + 1) debug(vi(E[i], E[i] + 2 * D + 1));
	rep(i, 0, D) {
		rep(j, 0, D) {
			if(E[i + l][j + l] - E[i][j + l] - E[i + l][j] + E[i][j] == lcnt) {
				int x = i + l, y = j + l;
				int nx = i + D, ny = j + D;
				if(F[nx][ny] - F[nx][y] - F[x][ny] + F[x][y] == 0) return true;
			}
		}
	}
	return false;
}

void solve() {
	cin >> N >> D;
	rep(i, 0, N) {
		int x, y; cin >> x >> y;
		S[x % D][y % D]++;
	}
	ll lv = 0, rv = 1000000000;
	while(rv - lv > 1) {
		ll m = (lv + rv) / 2;
		if(ok(m)) rv = m;
		else lv = m;
	}
	cout << rv - 1 << "\n";
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
	srand((unsigned int)time(NULL));
#ifdef LOCAL
	//freopen("in.txt", "wt", stdout); //for tester
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}