CodeForces 554B. Ohana Cleans Up【思维】

B. Ohana Cleans Up

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.

Return the maximum number of rows that she can make completely clean.

Input

The first line of input will be a single integer n (1 ≤ n ≤ 100).

The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.

Output

The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.

Examples

input

4
0101
1000
1111
0101

output

2

input

3
111
111
111

output

3

Note

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.

In the second sample, everything is already clean, so Ohana doesn't need to do anything.

解题思路：

其实就是找到最大的相同的行，有多少个。

#include<stdio.h>
#include<string.h>
#include<algorithm> 
using namespace std;
char map[1000][1000];
bool xx(int a,int b,int n)
{
	int i,j;
	for(i=0;i<n;i++)
	{
		if(map[a][i]!=map[b][i])
		return 0; 
	}
	return 1;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j;
		for(i=0;i<n;i++)
		{
				scanf("%s",map[i]);
		}
		int ans=0;
		for(i=0;i<n;i++)
		{
			int wc=0;
			for(j=0;j<n;j++)
			{
				if(xx(i,j,n)==1)
				{
					wc++;
				}
			}
			ans=max(ans,wc);
		}
		printf("%d\n",ans);
	}
	return 0;
}