Problem:
Score : 700 points

Problem Statement

Takahashi is hosting an sports meet.
There are N people who will participate. These people are conveniently numbered 1 through N.
Also, there are M options of sports for this event. These sports are numbered 1 through M.
Among these options, Takahashi will select one or more sports (possibly all) to be played in the event.

Takahashi knows that Person i's j-th favorite sport is Sport A_{ij}.
Each person will only participate in his/her most favorite sport among the ones that are actually played in the event, and will not participate in the other sports.

Takahashi is worried that one of the sports will attract too many people.
Therefore, he would like to carefully select sports to be played so that the number of the participants in the sport with the largest number of participants is minimized.
Find the minimum possible number of the participants in the sport with the largest number of participants.

Constraints

1 \leq N \leq 300

1 \leq M \leq 300

A_{i1} , A_{i2} , ... , A_{iM} is a permutation of the integers from 1 to M.

Input

Input is given from Standard Input in the following format:

N M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
:
A_{N1} A_{N2} ... A_{NM}

Output

Print the minimum possible number of the participants in the sport with the largest number of participants.

Sample Input 1

4 5
5 1 3 4 2
2 5 3 1 4
2 3 1 4 5
2 5 4 3 1

Sample Output 1

2

Assume that Sports 1, 3 and 4 are selected to be played. In this case, Person 1 will participate in Sport 1, Person 2 in Sport 3, Person 3 in Sport 3 and Person 4 in Sport 4.
Here, the sport with the largest number of participants is Sport 3, with two participants.
There is no way to reduce the number of participants in the sport with the largest number of participants to 1. Therefore, the answer is 2.

Sample Input 2

3 3
2 1 3
2 1 3
2 1 3

Sample Output 2

3

Since all the people have the same taste in sports, there will be a sport with three participants, no matter what sports are selected.
Therefore, the answer is 3.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[][] A = new int[N][M];
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < M; j++) {
				A[i][j] = sc.nextInt();
			}
		}

		boolean[] mask = new boolean[M];
		Arrays.fill(mask, true);
		int min = 99999999;
		for (int i = 0; i < M; i++) {
			int[] sum = new int[M];
			for (int j = 0; j < N; j++) {
				int index = getMaxIndex(A[j], mask);
				sum[index]++;
			}

			int maxIndex = getMaxIndex(sum, mask);
			min = Math.min(min, sum[maxIndex]);
			debug(sum, mask, maxIndex, min);
			mask[maxIndex] = false;
		}

		System.out.println(min);
	}

	private static int getMaxIndex(int[] a, boolean[] mask) {
		int max = -1;
		int index = -1;
		for (int i = 0; i < a.length; i++) {
			if (mask[i] && max < a[i]) {
				max = a[i];
				index = i;
			}
		}

		return index;
	}

	private static boolean isDebug = System.getProperty("sun.desktop") != null;

	private static void debug(Object... o) {
		if (isDebug) {
			System.err.println(Arrays.deepToString(o));
		}
	}

	public static class Scanner {
		private BufferedInputStream inputStream;
		private StringBuilder buffer = new StringBuilder();

		public Scanner(InputStream in) {
			inputStream = new BufferedInputStream(in);
		}

		public int nextInt() throws IOException {
			int num = 0;
			int sign = 1;

			int read = skip();
			if (read == '-') {
				sign = -1;
				read = inputStream.read();
			}

			do {
				num = num * 10 + sign * (read - 0x30);
			} while ((read = inputStream.read()) > 0x20);

			return num;
		}

		public void fill(int[] a) throws IOException {
			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
			}
		}

		public void fill(int[] a, int[] b) throws IOException {
			if (a.length != b.length) {
				throw new IllegalArgumentException();
			}

			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
				b[i] = nextInt();
			}
		}

		public long nextLong() throws IOException {
			long num = 0;
			int sign = 1;

			int read = skip();
			if (read == '-') {
				sign = -1;
				read = inputStream.read();
			}

			do {
				num = num * 10 + sign * (read - 0x30);
			} while ((read = inputStream.read()) > 0x20);

			return num;
		}

		public void fill(long[] a) throws IOException {
			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
			}
		}

		public void fill(long[] a, long[] b) throws IOException {
			if (a.length != b.length) {
				throw new IllegalArgumentException();
			}

			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
				b[i] = nextLong();
			}
		}

		public long[] nextLong(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++) {
				array[i] = nextLong();
			}

			return array;
		}

		public String next() throws IOException {
			int read = skip();
			do {
				buffer.append((char) read);
			} while ((read = inputStream.read()) > 0x20);

			String text = buffer.toString();
			buffer.delete(0, buffer.length());

			return text;
		}

		private int skip() throws IOException {
			int read;
			while ((read = inputStream.read()) <= 0x20)
				;

			return read;
		}
	}
}