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>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod

int usqrt(int x){
    FOR(i, 1, x + 1) if(i * i >= x) return i;
    return -1;
}

signed main(){

    int N, D; cin >> N >> D;
    map<Pii, int> cnt;
    REP(i, N){
        int x, y;
        cin >> x >> y;
        cnt[Pii(x % D, y % D)] += 1;
    }

    int S0 = -1, S1 = bit(20) - 1;
    mat a(2 * D, vec(2 * D, 0));
    while(S1 - S0 > 1){
        int S = (S0 + S1) / 2, T = S / D, r = S % D;
        REP(i, 2 * D) fill(all(a[i]), 0);
        for(PPi t: cnt){
            int x = t.f.f, y = t.f.s, n = t.s;
            if(T * T >= n){
                a[D - 1][D - 1]++;
                a[2 * D - 1][D - 1]--;
                a[D - 1][2 * D - 1]--;
                a[2 * D - 1][2 * D - 1]++;
            }else if(T * (T + 1) >= n){
                a[D - 1][D - 1]++;
                a[2 * D - 1][D - 1]--;
                a[D - 1][2 * D - 1]--;
                a[2 * D - 1][2 * D - 1]++;

                a[x][y]--;
                a[x - r + D - 1][y]++;
                a[x][y - r + D - 1]++;
                a[x - r + D - 1][y - r + D - 1]--;
            }else if((T + 1) * (T + 1) >= n){
                a[x - r + D - 1][y - r + D - 1]++;
                a[x + D][y - r + D - 1]--;
                a[x - r + D - 1][y + D]--;
                a[x + D][y + D]++;
            }
        }
        REP(i, 2 * D) REP(j, 2 * D - 1) a[i][j + 1] += a[i][j];
        REP(j, 2 * D) REP(i, 2 * D - 1) a[i + 1][j] += a[i][j];

        bool f = false;
        REP(i, D) REP(j, D) if(a[i][j] + a[i + D][j] + a[i][j + D] + a[i + D][j + D] >= SZ(cnt)) f = true;
        if(f) S1 = S;
        else S0 = S;
    }
    Out(S1);

    return 0;
}