CF 337A(Puzzles-暴力枚举)

A. Puzzles

time limit per test

memory limit per test

input

output

n

m puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of f₁ pieces, the second one consists of f₂

A be the number of pieces in the largest puzzle that the teacher buys and B be the number of pieces in the smallest such puzzle. She wants to choose such n puzzles that A - B is minimum possible. Help the teacher and find the least possible value of A - B.

Input

n and m (2 ≤ n ≤ m ≤ 50). The second line contains m space-separated integersf₁, f₂, ..., f_m (4 ≤ f_i ≤ 1000) — the quantities of pieces in the puzzles sold in the shop.

Output

Print a single integer — the least possible difference the teacher can obtain.

Sample test(s)

input

4 6 10 12 10 7 5 22

output

5

Note

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

排序，暴力找前后差m-1个的差即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000009)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+abs(a-b)/F*F+F)%F;}
typedef long long ll;
ll n,m,k;
ll pow2(ll a,int b)
{
  if (b==0) return 1;
  if (b==1) return a%F;
  ll p=pow2(a,b/2);
  p=p*p%F;
  if (b%2) p=p*a%F;
  return p;
}
int main()
{
//  freopen("Quiz.in","r",stdin);
//  freopen(".out","w",stdout);
//  For(i,100) cout<<pow2(2,i)<<' ';
  
  cin>>n>>m>>k;
  ll m1=m;m=n-m;
  int maxk=n/k;
  if (m>=maxk) {cout<<m1%F<<endl;return 0;}
  else m=maxk-m;
  
  ll am=sub((4*pow2(2,m-1))%F,2);
  
  ll ans=add(am*k%F,m1-k*m);
  cout<<ans%F<<endl;
  
  
  
  return 0;
}