A. Marks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook.

Overall the group has n students. They received marks for m subjects. Each student got a mark from 1 to 9 (inclusive) for each subject.

Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at.

Your task is to find the number of successful students in the group.

Input

The first input line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of students and the number of subjects, correspondingly. Next n lines each containing m characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.

Output

Print the single number — the number of successful students in the given group.

Sample test(s)

input

3 3
223
232
112

output

2

input

3 5
91728
11828
11111

output

3

Note

In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject.

In the second sample test each student is the best at at least one subject.

解题说明：此题就是判断一列数中最大的数字的个数

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

int main()
{
	int n,m;
	char a[101][102];
	int b[101];
	int max;
	int count;
	int i,j;
	scanf("%d %d",&n,&m);
	for(i=0;i<n;i++)
	{
		b[i]=0;
	}
	for(i=0;i<n;i++)
	{
		scanf("%s",&a[i]);
	}
	for(j=0;j<m;j++)
	{
		max=a[0][j]-'0';
		for(i=0;i<n;i++)
		{
			if(a[i][j]-'0'>max)
			{
				max=a[i][j]-'0';
			}
		}
		for(i=0;i<n;i++)
		{
			if(a[i][j]-'0'==max)
			{
				b[i]=1;
			}
		}
	}
	count=0;
	for(i=0;i<n;i++)
	{
		if(b[i]==1)
		{
			count++;
		}
	}
	printf("%d\n",count);

	return 0;
}