CodeForces - 124B-Permutations(DFS)

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本文探讨了一种算法问题，即如何通过重新排列一组k位数的整数中的数字，使得这些整数之间的最大值和最小值之差达到最小。文章详细介绍了输入输出格式，并提供了示例说明。该问题涉及深度优先搜索和回溯算法。

You are given n k-digit integers. You have to rearrange the digits in the integers so that the difference between the largest and the smallest number was minimum. Digits should be rearranged by the same rule in all integers.

Input

The first line contains integers n and k — the number and digit capacity of numbers correspondingly (1 ≤ n, k ≤ 8). Next n lines contain k-digit positive integers. Leading zeroes are allowed both in the initial integers and the integers resulting from the rearranging of digits.

Output

Print a single number: the minimally possible difference between the largest and the smallest number after the digits are rearranged in all integers by the same rule.

Examples

Input

Output

Input

Output

Input

Output

Note

In the first sample, if we rearrange the digits in numbers as (3,1,4,2), then the 2-nd and the 4-th numbers will equal 5237 and 2537 correspondingly (they will be maximum and minimum for such order of digits).

In the second sample, if we swap the second digits and the first ones, we get integers 100, 99 and 102.

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
char s[100][100],str[100];
int vis[100],k,cha,b[100],n;
#define inf 99999999
void dfs(int wei)
{
	int i,j,minx=inf,maxx=-inf,num;
	if(wei==k+1){
	
		for(i=1;i<=n;i++){
			num=0;
			for(j=1;j<=k;j++){
				num=num*10+s[i][b[j]]-'0';
			}
			maxx=max(maxx,num);
			minx=min(minx,num);
		}
		cha=min(cha,maxx-minx);//return;
	}
	else{
		 for(i=1;i<=k;i++){
		   if(!vis[i]){
			  vis[i]=1;
			  b[wei]=i;
			  dfs(wei+1);
			  vis[i]=0;
		   }
	    }
	}
}
 
 
int main()
{
	int m,i,j,t,num;
	char c;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		for(i=1;i<=n;i++){
			scanf("%s",s[i]+1);
		}
		cha=inf;
		if(n==1){
			printf("0\n");continue;
		}
		memset(vis,0,sizeof(vis));
		dfs(1);
		printf("%d\n",cha);
	}
	return 0;
}