Codeforces Round #438

C. Qualification Rounds

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of nproblems, and they want to select any non-empty subset of it as a problemset.

k experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.

Determine if Snark and Philip can make an interesting problemset!

Input

The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 4) — the number of problems and the number of experienced teams.

Each of the next n lines contains k integers, each equal to 0 or 1. The j-th number in the i-th line is 1 if j-th team knows i-th problem and 0 otherwise.

Output

Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

Examples

input

5 31 0 11 1 01 0 01 0 01 0 0

output

NO

input

3 21 01 10 1

output

YES

Note

In the first example you can't make any interesting problemset, because the first team knows all problems.

In the second example you can choose the first and the third problems.

题意：

给你几个问题，和每个问题对于k个人他们所了不了解。

让你判断是否存在挑出任意几个问题，能让这k个人最多只了解这些问题的一半。

POINT：

答案必为挑2个问题组合，状态压缩一下。

#include <iostream>#include <string.h>#include <stdio.h>#include <map>#include <algorithm>using namespace std;#define LL long longconst int maxn = 20;map<int,int>mp;int main(){    int n,k;    scanf("%d %d",&n,&k);    int cnt[maxn];    int num=0;    for(int i=1;i<=n;i++){        int now=0;        for(int j=0;j<k;j++){            int a;            scanf("%d",&a);            if(a==1)                now=now|(1<<j);        }        if(mp[now]==0){            mp[now]=1;            cnt[++num]=now;        }    }    for(int i=1;i<=num;i++){        for(int j=1;j<=num;j++){            int now=0;            for(int s=0;s<=k;s++){                now=now|((cnt[i]>>s)&(cnt[j]>>s));            }            if(now==0){                printf("YES\n");                return 0;            }        }    }    printf("NO\n");}