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: Wrong Answer)

#include<algorithm>
#include<iomanip>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<vector>
#include<bitset>
#include<cmath>
#define FOR(i,a,b) for(int i=(a); i<int(b); ++i)
#define RFOR(i,a,b) for(int i=(b)-1; i>=int(a); --i)
#define rep(i,n) FOR(i,0,n)
#define rep1(i,n) FOR(i,1,int(n)+1)
#define rrep(i, n) RFOR(i, 0, n)
#define rrep1(i, n) RFOR(i, 1, int(n) + 1)
#define all(c) begin(c), end(c)
#define al(d) d.begin(),d.end()
#define fill(n) std::setfill('0')<<std::right<<std::setw(n)
template <typename T>
using vec = std::vector<T>;
using vi = vec<int>;
using vvi = vec<vi>;
using pii = std::pair<int, int>;
using ll = long long;
using ld = long double;
using namespace std;
const int MOD = 1000000007;

template <typename T>
void putv(vector<T>& V){
	// cout << "The elements in the vector are: " << endl;
	for(auto x: V)
		cout << x << " ";
	cout << endl;
}
template <class T>
vector<T> getv(int n){
	vector<T> vec;
	rep(i,n){
		T input;
		cin >> input;
		vec.emplace_back(input);
	}
	return vec;
}

using AdjList = vector<vector<int>>;
using Weight = int;
using Flow = int;
struct Edge {
	int src, dst;
	Weight weight;
	Flow cap;
	Edge() : src(0), dst(0), weight(0) {}
	Edge(int s, int d, Weight w) : src(s), dst(d), weight(w) {}
};
using Edges = std::vector<Edge>;
using Graph = std::vector<Edges>;
using Array = std::vector<Weight>;
using Matrix = std::vector<Array>;
void add_edge(Graph &g, int a, int b, Weight w = 1) {
	g[a].emplace_back(a, b, w);
	g[b].emplace_back(b, a, w);
}
void add_arc(Graph &g, int a, int b, Weight w = 1) { g[a].emplace_back(a, b, w); }
// Recieve edge(=(int,int)) as an adj list.
AdjList getadj(int N, int M){
	// Initialize all lists of the graph as a empty vector.
	vector<int> init;
	vector<vector<int>> graph(N,init);
	rep(i,M){
		int a, b;
		cin >> a >> b;
		graph[a-1].emplace_back(b-1);
		graph[b-1].emplace_back(a-1);
	}
// rep(i,N)
//  putv(graph[i]);
	return graph;
}

ll Manhattan_dist(ll x1, ll y1, ll x2, ll y2){
	return abs(x1-x2) + abs(y1-y2);
}
void out(bool ans){
	if(ans == true) cout << "yes" << endl;
	else cout << "no" << endl;
}
void Out(bool ans){
	if(ans == true) cout << "Yes" << endl;
	else cout << "No" << endl;
}
void OUT(bool ans){
	if(ans == true) cout << "YES" << endl;
	else cout << "NO" << endl;
}
ll gcd(const ll a, const ll b){
	return (b==0) ? a : gcd(b,a%b);
}
ll lcm(const ll a, const ll b){
	return a*(b/gcd(a,b));
}
// イテレーティブな方法で適宜組んだ方が時間量領域量ともにセーブできる
// ll lcmv(vector<ll> vec){
// 	int t = vec.size();
// 	if(t==2)
// 		return lcm(vec[0],vec[1]);
// 	ll v = vec[t-1];
// 	vec.pop_back();
// 	return lcm(lcmv(vec),v);
// }
// ll gcdv(vector<ll> vec){
// 	int t = vec.size();
// 	if(t==1)
// 		return vec[0];
// 	if(t==2)
// 		return gcd(vec[0],vec[1]);
// 	ll v = vec[t-1];
// 	vec.pop_back();
// 	return gcd(gcdv(vec),v);
// }

// $\min\{s\in\mathbb{N}:p^s|n\}$
// nをpで何回割れるか
ll hmt_p_divide_n(ll p, ll n){
	ll s = 0;
	while(n%p == 0){
		n /= p;
		s++;
	}
	return s;
}

int main(){
	int N,K;
	cin >>N>>K;

	vi h =getv<int>(N);
	sort(all(h));
	putv(h);
	int min_dist = 1e9;
	rep(i,N-K+1){
		if(h[i+K-1]-h[i] < min_dist) cout << h[i+K-1]-h[i] << "の方が小さいから更新" << endl;
		min_dist = min(min_dist, h[i+K-1]-h[i]);
	}
	cout << min_dist << endl;
}