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)

#define _CRT_SECURE_NO_WARNINGS

/* include ***********************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

/* define *************************/
//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define FOR(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)
#define RFOR(i,s,n) for(int i=(int)(s)-1,i##_len=(int)(n);i>=i##_len;i--)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTL(d) printf("%lld\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9
#define MOD 1000000007

typedef long long ll;
typedef unsigned long long ull;

int p[100000];
int y[100000];
int z[100000];
int ans[100000];

int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int  Max(int  a, int  b) {
	return (a) > (b) ? (a) : (b);
}
void Swap(int *a, int *b) {
	int tmp = *a;
	*a = *b;
	*b = tmp;
}void hSwap(int x[], int i, int j)
{
	int temp;

	temp = x[i];
	x[i] = x[j];
	x[j] = temp;
}

void ShowData(int x[], int left, int right)
{
	int i;
	for (i = left; i <= right; i++)printf("%d  ", x[i]);
	printf("\n");
}
void QSort(int x[], int left, int right, int n) {
	int i, j;	//左端,右端
	int pivot;	//軸

	i = left;
	j = right;
	pivot = x[(left + right) / 2];

	while (1) {
		if (n > 0) {				//n>0なら昇順、n<=0なら降順
			while ((x[i] < pivot) && (i <= right))i++;	//軸値より大きい要素
			while ((pivot < x[j]) && (i <= right))j--;	//軸値より小さい要素
		}
		else {
			while ((x[i] > pivot) && (i <= right))i++;	//軸値より小さい要素
			while ((pivot > x[j]) && (i <= right))j--;	//軸値より大きい要素
		}
		if (i >= j)	break;
		hSwap(x, i, j);
		i++; j--;
	}
	//ShowData(x, left, right);

	if (left < i - 1)QSort(x, left, i - 1, n);
	if (j + 1 <  right)QSort(x, j + 1, right, n);
}

int gcd(int a, int b) {
	if (b == 0)return a;
	int tmp = a%b;
	a = b;
	b = tmp;
	return gcd(a, b);
}


int main() {
	int n, k;
	scanf("%d%d", &n, &k);

	int h[100000];
	REP(i, n) {
		scanf("%d", &h[i]);
	}
	QSort(h, 0, n - 1, 1);
	int min = INF;
	REP(i, n - k + 1) {
		min = Min(min, h[i + k - 1] - h[i]);
	}

	PRINTD(min);
}