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在 n 道題目中挑選一些使得所有人對題目的掌握情況不超過一半。

als main equal 超過一半 %d each amp preview 掌握

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
n 道題目中挑選一些使得所有人對題目的掌握情況不超過一半。
#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");
}

在 n 道題目中挑選一些使得所有人對題目的掌握情況不超過一半。