Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined)C

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 0otherwise.

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：有n道题目，k个参加队伍，对于每个题目参加的队伍可能知道这个题目也可能不知道这个题目，你想要找到一个关于这n道题的非空子集，每个参加队伍知道的题目不超过总数的一半。问是否能找到这么个子集。

思路：这题还挺有意思的，首先你会先到如果存在一道题是全部队伍都不知道，那么答案肯定是yes，否则不存在选一道题的情况，然后如果选两道题，那么答案为yes的最坏的情况就是每个队伍都知道一道题，然后你会发现如果两道题解决不了的话三道题四道题也同样解决不了，所以最坏的就是选两道题，所以每次判一下两道题能不能满足条件就好了。下面给代码：

#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>typedef long long LL;using namespace std;#define inf 0x3f3f3f3f#define maxn 20typedef long long LL;int vis[maxn];int main(){int n, k;scanf("%d%d", &n, &k);bool jud = false;while (n--){int status = 0;for (int i = 0; i < k; i++){int x;scanf("%d", &x);if (x)status |= 1 << i;}vis[status] = 1;for (int i = 0; i < (1 << k); i++){if (status&i)continue;if (vis[i])jud = true;}}if (jud)puts("YES");elseputs("NO");}