codeforces463D

Gargari and Permutations

CodeForces - 463D

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].

sol:求m个序列的lcs，似乎看上去很难得样子，实际上想到dp状态就不难了，dp[i]表示以数字 i 结尾的m个序列的lcs长度，然后判断一下是否所有数字 j 都在 i 之前就可以转移了

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+''); return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,m,a[][N],Pos[][N];
int dp[N];
int main()
{
    int i,j,k;
    R(n); R(m);
    for(i=;i<=m;i++)
    {
        for(j=;j<=n;j++)
        {
            R(a[i][j]); Pos[i][a[i][j]]=j;
        }
    }
    dp[a[][]]=;
    for(i=;i<=n;i++)
    {
        for(j=;j<i;j++)
        {
            bool Flag=;
            for(k=;k<=m&&Flag;k++) if(Pos[k][a[][j]]>Pos[k][a[][i]]) Flag=;
            if(Flag) dp[a[][i]]=max(dp[a[][i]],dp[a[][j]]+);
        }
        if(!dp[a[][i]]) dp[a[][i]]=;
    }
    int ans=;
    for(i=;i<=n;i++) ans=max(ans,dp[i]);
    Wl(ans);
    return ;
}
/*
Input
4 3
1 4 2 3
4 1 2 3
1 2 4 3
Output
3
*/