Educational Codeforces Round 44 (Rated for Div. 2) B. Switches and Lamps

B. Switches and Lamps

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n switches and m lamps. The i-th switch turns on some subset of the lamps. This information is given as the matrix aconsisting of n rows and m columns where ai, j = 1 if the i-th switch turns on the j-th lamp and ai, j = 0 if the i-th switch is not connected to the j-th lamp.

Initially all m lamps are turned off.

Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.

It is guaranteed that if you push all n switches then all m lamps will be turned on.

Your think that you have too many switches and you would like to ignore one of them.

Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other n - 1 switches then all the m lamps will be turned on.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of the switches and the number of the lamps.

The following n lines contain m characters each. The character ai, j is equal to '1' if the i-th switch turns on the j-th lamp and '0' otherwise.

It is guaranteed that if you press all n switches all m lamps will be turned on.

Output

Print "YES" if there is a switch that if you will ignore it and press all the other n - 1 switches then all m lamps will be turned on. Print "NO" if there is no such switch.

Examples

input

Copy

4 5
10101
01000
00111
10000

output

Copy

YES

input

Copy

4 5
10100
01000
00110
00101

output

Copy

NO

题意：给一个n行m列的矩阵，n表示n个开关,m表示m盏灯

矩阵中1表示该开关能控制该灯，为0表示不能，问能否用n-1个开关控制m盏灯，若能输出YES，否则输出NO

题解：

就看每盏灯是否能被单独控制，即少了这个开关不行，如果每个开关都单独控制一盏灯，就NO，其他就YES

#include <bits/stdc++.h>
#define maxn 2005 
using namespace std;
char a[maxn][maxn];
int v[maxn];
int n,s,t,m; 
int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++)scanf("%s",a[i]);
     for(int j=0;j<m;j++)
     {
      s=0;
      for(int i=0;i<n;i++)
        if(a[i][j]-'0'==1)s++,t=i;
        
         if(s==1) v[t]=1;
}
   for(int i=0;i<n;i++)
    if(!v[i]) { printf("YES\n"); return 0;}
    printf("NO\n");
return 0;
}