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=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=1005;
int n,m,a[10][N],Pos[10][N];
int dp[N];
int main()
{
    int i,j,k;
    R(n); R(m);
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            R(a[i][j]); Pos[i][a[i][j]]=j;
        }
    }
    dp[a[1][1]]=1;
    for(i=2;i<=n;i++)
    {
        for(j=1;j<i;j++)
        {
            bool Flag=1;
            for(k=2;k<=m&&Flag;k++) if(Pos[k][a[1][j]]>Pos[k][a[1][i]]) Flag=0;
            if(Flag) dp[a[1][i]]=max(dp[a[1][i]],dp[a[1][j]]+1);
        }
        if(!dp[a[1][i]]) dp[a[1][i]]=1;
    }
    int ans=1;
    for(i=1;i<=n;i++) ans=max(ans,dp[i]);
    Wl(ans);
    return 0;
}
/*
Input
4 3
1 4 2 3
4 1 2 3
1 2 4 3
Output
3
*/