Description：

Solution：

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<queue>

#include<vector>

#include<cctype>

#include<cstring>

using namespace std;

int n,m;

#define MAXN 100010

typedef long long ll;

priority_queue<ll,vector<ll>,greater<ll> > q;

struct node

{

	int lc,rc;

	int d[2],mx[2],mn[2];

	int& operator [](int x){return d[x];}

}p[MAXN],t[MAXN];

int ptr = 0;

int newnode(){return ++ptr;}

int root;

int cmpd;

bool operator < (node a,node b){return a[cmpd] < b[cmpd];}

void updata(int rt)

{

	for(int i = 0;i <= 1;++i)

	{

		if(t[rt].lc != 0)

		{

			t[rt].mn[i] = min(t[rt].mn[i],t[t[rt].lc].mn[i]);

			t[rt].mx[i] = max(t[rt].mx[i],t[t[rt].lc].mx[i]);

		}

		if(t[rt].rc != 0)

		{

			t[rt].mn[i] = min(t[rt].mn[i],t[t[rt].rc].mn[i]);

			t[rt].mx[i] = max(t[rt].mx[i],t[t[rt].rc].mx[i]);

		}

	}

	return;

}

void build(int &rt,int l,int r,int type)

{

	if(l > r)return;

	rt = newnode();

	int mid = ((l + r) >> 1);

	cmpd = type;

	nth_element(p + l,p + mid,p + r + 1);

	t[rt] = p[mid];

	for(int i = 0;i <= 1;++i)t[rt].mn[i] = t[rt].mx[i] = t[rt].d[i];

	build(t[rt].lc,l,mid - 1,type ^ 1);

	build(t[rt].rc,mid + 1,r,type ^ 1);

	updata(rt);

	return;

}

ll sqr(int x){return 1ll * x * x;}

ll dis(node a,node b){return sqr(a[0] - b[0]) + sqr(a[1] - b[1]);}

ll calc(node a,node b)

{

	return max(sqr(a.mn[0] - b[0]),sqr(a.mx[0] - b[0])) + max(sqr(a.mn[1] - b[1]),sqr(a.mx[1] - b[1]));

}

#define INF 0x3f3f3f3f3f3f3f3f

void query(int rt,node k)

{

	ll d,dl = -INF,dr = -INF;

	d = dis(t[rt],k);

	if(d > q.top()){q.pop();q.push(d);}

	if(t[rt].lc)dl = calc(t[t[rt].lc],k);

	if(t[rt].rc)dr = calc(t[t[rt].rc],k);

	if(dl > dr)

	{

		if(dl > q.top())query(t[rt].lc,k);

		if(dr > q.top())query(t[rt].rc,k);

	}

	else

	{

		if(dr > q.top())query(t[rt].rc,k);

		if(dl > q.top())query(t[rt].lc,k);

	}

	return;

}

int main()

{

	scanf("%d%d",&n,&m);

	for(int i = 1;i <= 2 * m;++i)q.push(0);

	for(int i = 1;i <= n;++i)

	{

		scanf("%d%d",&p[i][0],&p[i][1]);

	}

	build(root,1,n,0);

	for(int i = 1;i <= n;++i)query(root,p[i]);

	cout << q.top() << endl;

	return 0;

}