Codeforce 124b--Full arrangement dfs--permutations

You are given nk-digit integers. Rearrange the digits in the integers so, the difference between the largest and the smallest number was MI Nimum. Digits should is rearranged by the same rule in all integers.

Input

The first line contains integers n and K -the number and digit capacity of numbers Correspondin Gly (1≤ n, k ≤8). Next n lines containk-digit positive integers. Leading zeroes is 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 is Rearranged in all integers by the same rule.

Sample Input

Input

6 4
5237
2753
7523
5723
5327
2537

Output

2700

Input

7 ·
010
909
012

Output

3

Input

9 {
50808
36603
37198
44911
29994
42543
50156

Output

20522

Hint

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 would equal 523 7 and 2537 correspondingly (they would 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, and 102.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int inf = 0x3f3f3f3f ; const int INF = -0x3f3f3f3f;char a[10][10];int b[10][10];int c[10];int vis[10];int ans[10];int N, k;int anser;void dfs (i       NT CNT) {if (cnt = = k+1) {memset (c,0,sizeof (c));               for (int i = 1, i <= N; i++) {for (int j = 1; j <= K; j + +) {C[i] = c[i]*10 + b[i][ans[j]];      }} sort (c + 1, C + n + 1);       printf ("%d%d\n", c[n],c[1]);      Anser = min (Anser, C[n]-c[1]);     return;            } for (int i = 1; I <= K; i++) {if (vis[i] = = 0) {ans[cnt] = i;            Vis[i] = 1;            DFS (CNT+1);        Vis[i] = 0;        }}}int Main () {while (~scanf ("%d%d", &n,&k)) {anser = inf;        for (int i = 1; I <= n; i++) scanf ("%s", &a[i]); for (int i = 1, i <= N; i++) {for (int j = 0;j < K; J + +) {b[i][j+1] = a[i][j]-' 0 ';           }}/* for (int i = 1, i <= N; i++) {for (int j = 1; j <= K; j + +)          printf ("%d", b[i][j]);        Puts ("");    }*/memset (vis,0,sizeof (VIS));    DFS (1);    printf ("%d\n", Anser); } return 0;}

　　

Codeforce 124b--Full arrangement dfs--permutations