Problem: You are given n arrays a_1, a_2, ..., a_n; each array consists of exactly m integers. We denote the y-th element of the x-th array as a_{x, y}.

You have to choose two arrays a_i and a_j (1 ≤ i, j ≤ n, it is possible that i = j). After that, you will obtain a new array b consisting of m integers, such that for every k ∈ [1, m] b_k = max(a_{i, k}, a_{j, k}).

Your goal is to choose i and j so that the value of min _{k = 1}^{m} b_k is maximum possible.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 8) — the number of arrays and the number of elements in each array, respectively.

Then n lines follow, the x-th line contains the array a_x represented by m integers a_{x, 1}, a_{x, 2}, ..., a_{x, m} (0 ≤ a_{x, y} ≤ 10^9).

Output

Print two integers i and j (1 ≤ i, j ≤ n, it is possible that i = j) — the indices of the two arrays you have to choose so that the value of min _{k = 1}^{m} b_k is maximum possible. If there are multiple answers, print any of them.

Example

Input


6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0


Output


1 5

Solution (in JAVA):

Solution:
//package EduRound80;

import java.util.Scanner;

/**
 * @author sguar <shugangcao@gmail.com>
 * welcome to kuaishou
 * Created on 2020-01-18
 */
public class D {
    public static void main(String [] args) {
        new D().run();
    }

    int [] a;
    int [] maxMin;
    int [] index;
    int [] nowMin;
    private void run() {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int m = cin.nextInt();
        a = new int[m];
        int cnt = (1 << m) - 1;
        maxMin = new int[cnt + 1];
        nowMin = new int[cnt + 1];
        index  = new int[cnt + 1];
        int ans = 0;
        int ansOfI = 0;
        int ansOfJ = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[j] = cin.nextInt();
            }

            for (int k = 0; k <= cnt; k++) {
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < m; j++) {
                    if (((k >> j) & 1) == 1) {
                        min = Math.min(min, a[j]);
                    }
                }
                nowMin[k] = min;
                if (Math.min(min, maxMin[cnt - k]) > ans) {
                    ans = Math.min(min, maxMin[cnt - k]);
                    ansOfI = index[cnt - k];
                    ansOfJ = i;
                }
            }
            for (int k = 0; k <= cnt; k++) {
                if (nowMin[k] > maxMin[k]) {
                    maxMin[k] = nowMin[k];
                    index[k] = i;
                }
            }
        }
        System.out.println((ansOfI + 1) + " " + (ansOfJ + 1));
    }
}