CodeForces

Discription

Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have found k permutations. Each of them consists of numbers 1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?

You can read about longest common subsequence there:https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Input

The first line contains two integers n and k (1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the next k lines contains integers 1, 2, ..., n in some order — description of the current permutation.

Output

Print the length of the longest common subsequence.

Examples

Input

4 3
1 4 2 3
4 1 2 3
1 2 4 3

Output

3

Note

The answer for the first test sample is subsequence [1, 2, 3].

   一开始看的时候有点懵，，，，多个串的最长公共子序列 我好像只会后缀自动机啊hhhhh，这个题字符集大小还蛮大的emmmm(事实上是我根本不想写SAM   hhh)。

    但是这个题有特殊性质啊，每个串都是一个排列，也就是说每个数在k个串里只有一种组合方式，所以我们直接在DAG上dp就好了。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1005;
int P[maxn][5],n,now,k,to[maxn*1000];
int ne[maxn*1000],hd[maxn],num,ans,F[maxn];
inline void add(int x,int y){ to[++num]=y,ne[num]=hd[x],hd[x]=num;}
void dp(int x){
	if(F[x]) return;
	for(int i=hd[x];i;i=ne[i]) dp(to[i]),F[x]=max(F[x],F[to[i]]);
	F[x]++;
}
int main(){
	scanf("%d%d",&n,&k);
	for(int i=0;i<k;i++)
	    for(int j=1;j<=n;j++) scanf("%d",&now),P[now][i]=j;
	for(int i=1;i<=n;i++)
	    for(int j=1,o;j<=n;j++){
	    	for(o=0;o<k;o++) if(P[i][o]>=P[j][o]) break;
	    	if(o==k) add(i,j);
		}
	
	for(int i=1;i<=n;i++) dp(i),ans=max(ans,F[i]);
	printf("%d
",ans);
	return 0;
}