Problem:
Score : 300 points

Problem Statement

In some other world, today is Christmas Eve.

There are N trees planted in Mr. Takaha's garden. The height of the i-th tree (1 \leq i \leq N) is h_i meters.

He decides to choose K trees from these trees and decorate them with electric lights. To make the scenery more beautiful, the heights of the decorated trees should be as close to each other as possible.

More specifically, let the height of the tallest decorated tree be h_{max} meters, and the height of the shortest decorated tree be h_{min} meters. The smaller the value h_{max} - h_{min} is, the better. What is the minimum possible value of h_{max} - h_{min}?

Constraints

2 \leq K < N \leq 10^5

1 \leq h_i \leq 10^9

h_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
h_1
h_2
:
h_N

Output

Print the minimum possible value of h_{max} - h_{min}.

Sample Input 1

5 3
10
15
11
14
12

Sample Output 1

2

If we decorate the first, third and fifth trees, h_{max} = 12, h_{min} = 10 so h_{max} - h_{min} = 2. This is optimal.

Sample Input 2

5 3
5
7
5
7
7

Sample Output 2

0

If we decorate the second, fourth and fifth trees, h_{max} = 7, h_{min} = 7 so h_{max} - h_{min} = 0. This is optimal.

There are not too many trees in these sample inputs, but note that there can be at most one hundred thousand trees (we just can't put a sample with a hundred thousand lines here).

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

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;

//using Graph = vector<vector<edge>>;
using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i, n)         for(int i = 0; i < (int)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define erep(i,container) for (auto i : container)
#define irep(i, n)        for(int i = n-1; i >= (int)0; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define all(x) (x).begin(),(x).end()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
#define case std::cout <<"Case #" <<qqq+1<<": "
#define RES(a,i,j) a.resize(i);rep(ii,i)a[ii].resize(j); 
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout<<a<<" "<<b<<" "<<c<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define ssum(a) accumulate(a,0ll) 
#define gre(n) greater<n>()

typedef long long ll;
//
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
//typedef long long T;
const int INF = 2000000000;
const double dINF = 1000000000;
const ll INF64 = 922337203685477580ll;
const ll MOD = 1000000007ll;
//const ll MOD = 67280421310721ll;
const dou pi=3.141592653589793;
//メモ
//二次元配列のresizeのしかた
//vv.resize(n);
//for(size_t i=0; i<n; i++){
// vv[i].resize(n);
//}

//メモ
//負の辺を持つ最短経路問題では到達不可経路でも初期化値より小さくなりうるので注意しよう！


int main(){
    int n,k;
    std::cin >> n>>k;
    VEC(int,a,n);
    sort(all(a));
    int ans=INF;
    rep(i,n-k+1){
        chmin(ans,a[i+k-1]-a[i]);
    }
    std::cout << ans << std::endl;
}