在 n 道题目中挑选一些使得所有人对题目的掌握情况不超过一半。

Posted 2020-10-12 Billyshuai

Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of n problems, 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 3
1 0 1
1 1 0
1 0 0
1 0 0
1 0 0

Output

NO

Input

3 2
1 0
1 1
0 1

Output

YES
在 n 道题目中挑选一些使得所有人对题目的掌握情况不超过一半。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool vis[22];
int main(){
    int n,k,x;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;++i) {
    int sum=0;
    for(int j=1;j<=k;++j) {
        scanf("%d",&x);
        if(x) sum+=1<<(j-1);
    }
    vis[sum]=1;
    }
    bool flag=0;
    for(int i=0;i<=15;++i) for(int j=0;j<=15;++j) if(((i&j)==0)&&vis[i]&&vis[j]) flag=1; 
    if(flag) puts("YES");else puts("NO");
}