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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/

int main()
{
    int N;
    lint D;
    cin >> N >> D;
    vector<vector<int>> cou(D, vector<int>(D));
    REP(_, N)
    {
        int x, y;
        cin >> x >> y;
        cou[x % D][y % D]++;
    }
    vector<vector<vector<vector<lint>>>> dp;
    ndarray(dp, 2, 2 * D + 1, 2 * D + 1, 2);

    REP(i, 2 * D) REP(j, 2 * D)
    {
        vector<lint> now{i, j};
        int n = 1;
        int c = cou[i % D][j % D];
        while (n * n < c) n++;
        bool rect = false;
        if (n * n - c >= n + 1) rect = true;
        lint lo = (n - rect - 1) * D;
        lint hi = (n - 1) * D;
        // dbg((vector<lint>{c, n, D, lo, hi}));

        vector<lint> hilo{hi, lo};

        REP(e, 2)
        {
            mmax(dp[0][i][j + 1][e], now[e] + (e == 1 ? D : 0) + hilo[e]);
            mmax(dp[0][i + 1][j + 1][e], now[e] + D + hilo[e]);
            mmax(dp[1][i + 1][j][e], now[e] + (e == 0 ? D : 0) + hilo[!e]);
            mmax(dp[1][i + 1][j + 1][e], now[e] + D + hilo[!e]);
        }
    }
    // dbg(dp[0]);
    lint ret = 1e18;
    FOR(i, D, 2 * D) FOR(j, D, 2 * D) REP(x, 2)
    {
        mmax(dp[0][i][j + 1][x], dp[0][i][j][x]);
        mmax(dp[1][i + 1][j][x], dp[1][i][j][x]);
        REP(d, 2) mmax(dp[d][i + 1][j + 1][x], dp[d][i][j][x]);
        lint tmp = 0;
        REP(d, 2) mmax(tmp, max(dp[d][i][j][0] - i, dp[d][i][j][1] - j));
        mmin(ret, tmp);
    }
    cout << ret << endl;
}