D. Gargari and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

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

output

Copy

3

Note

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

补题dp：

题意：求k个长度为n的最长公共子序列

思路1：保存每个数在各自串的位置，因为结果是第1个串中的某个可能，所以我们枚举第1个串的可能，然后检查如果一个以a[j]为结束的最长公共子序列成立的情况是，对于每个串的a[i]都在a[j]的前面，那么就有dp[j] = max(dp[j], dp[i]+1)

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int b[10][N],a[10][N],dp[N];
int n,k;
bool valid(int x,int y)
{
	for(int i=2;i<=k;i++)
		if(b[i][x]>b[i][y]) return 0;
	return 1;
}
int main()
{
	cin>>n>>k;
	for(int i=1;i<=k;i++)
	for(int j=1;j<=n;j++)
	{
		scanf("%d",&a[i][j]);
		b[i][a[i][j]]=j;
	}
	for(int i=1;i<=n;i++) dp[i]=1;
	for(int i=1;i<n;i++)
	for(int j=i+1;j<=n;j++)
		if(valid(a[1][i],a[1][j]))
			dp[j]=max(dp[j],dp[i]+1);
	int ans=0;
	for(int i=1;i<=n;i++) ans=max(ans,dp[i]);
	cout<<ans;
}