codeforces 868c（二进制）

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

题意：输入一个n一个k分别代表题目数和队伍数，接下来n行每行k个数代表每道题每个队伍是否知道，判断能否找到这样的情况选出其中几题，每队都只知道至多一半的题。

思路：如果能选出几题满足，那么必定可以选出2题的情况满足。这样使用2进制表示每道题几个队伍的知道情况，这样这道题的情况就变成了一个小于16的数，知道找到两道题的情况&运算为0即可。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;bool data[16];int main(){int n,k;int x;while(~scanf("%d%d",&n,&k)){for(int i=0;i<n;i++){int cnt=0;for(int j=0;j<k;j++){scanf("%d",&x);cnt*=2;cnt+=x;}data[cnt]=1;}bool flag=false;for(int i=0;i<16;i++){int j;for(j=0;j<16;j++){if(((i&j)==0)&&data[i]&&data[j]){flag=true;break;}}if(j<16)break;}if(flag)printf("YES\n");else printf("NO\n");}}