A. Opponents

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

input

2 2
10
00

output

2

input

4 1
0100

output

1

input

4 5
110111110110
10111111

output

2

Note

In the first and the second samples, Arya will beat all present opponents each of the d days.

In the third sample, Arya will beat his opponents on days 1, 3 and 4 and
his opponents will beat him on days 2 and 5.
Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.

解题说明：此题是统计连续出现不全为1的字符串的最大次数，遍历判断字符串，判断字符串中是否不全为1.

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include <map>
using namespace std;

int main( )
{
int n,d,i,j,k=0,c=0,l=0,max=0;
char a[101];
scanf("%d %d",&n,&d);
for(i=0;i<d;i++)
{
c=0;
scanf("%s",a);
l=strlen(a);
for(j=0;j<l;j++)
{
if(a[j]=='1')
{
c+=1;
}
}
if(c!=l)
{
k+=1;
}
if(k>max)
{
max=k;
}
if(c==l)
{
k=0;
}
}
printf("%d\n",max);
return 0;
}