D - Opponents

Description

Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya.

For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents.

Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.

Input

The first line of the input contains two integers n and d (1 ≤ n, d ≤ 100) — the number of opponents and the number of days, respectively.

The i-th of the following d lines contains a string of length n consisting of characters '0' and '1'. The j-th character of this string is '0' if the j-th opponent is going to be absent on the i-th day.

Output

Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.

Sample Input

Input

2 2
10
00

Output

2

Input

4 1
0100

Output

1

Input

4 5
1101
1111
0110
1011
1111

Output

2

题意：

Arya与n个对手每天打一架，当这一天n个对手全部来时Arya就输了，否则就是Arya赢，求最大连胜天数。

附AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
 
int main( )
{
    int n,d,i,j,k=,c=,l=,max=;
    char a[];
    scanf("%d %d",&n,&d);
    for(i=;i<d;i++)
    {
        c=;
        scanf("%s",a);
        l=strlen(a);
        for(j=;j<l;j++)
        {
            if(a[j]=='')
            {
                c+=;
            }
        }
        if(c!=l)
        {
            k+=;
        }
        if(k>max)
        {
            max=k;
        }
        if(c==l)
        {
            k=;
        }
    }
    printf("%d\n",max);
    return ;
}