A. Marks

time limit per test

1 second

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 studentsuccessful, 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 from1 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 3223232112

output

2

input

3 5917281182811111

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.

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define MAXN 102
using namespace std;
char score[MAXN][MAXN];
int stu[MAXN] = {0};
int main()
{
int n, m, Max;
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i)
{
scanf("%s", &score[i]);
}
for(int j = 0; j < m; ++j)
{

Max = score[0][j] - '0';

for(int i = 0; i < n; ++i)
{
if(score[i][j] - '0' > Max)
{
Max = score[i][j] - '0';
}
}


for(int i = 0; i < n; ++i)
{
if(score[i][j] - '0' == Max)
{
stu[i] = 1;
}
}
}

int count=0;
for(int i = 0; i < n; ++i)
{
if(stu[i] == 1)
{
count++;
}
}
printf("%d\n", count);
return 0;
}