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:
import java.util.*;

public class Minimax {
    final static long MAX_VAL = Math.round(1e9);
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long max = 0;
        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();
                if (a[i][j] > max) {
                    max = a[i][j];
                }
            }
        }
        long min = 0;

        int p = 0, q = 0;
        long allOnes = (1 << m) - 1;
        while (max > min) {
            long t = (max + min) / 2 + 1;
            boolean feasible = false;
            Map<Long, Integer> binarized = new HashMap<>(256);
            for (int i = 0; i < n; i ++) {
                long b = 0;
                for (int j = 0; j < m; j++) {
                    if (a[i][j] >= t) {
                        b += 1 << j;
                    }
                }
                if (!binarized.containsKey(b)) {
                    binarized.put(b, i);
                }
            }

            out:
            for (long b1 : binarized.keySet()) {
                for (long b2 : binarized.keySet()) {
                    if ((b1 | b2) == allOnes) {
                        feasible = true;
                        p = binarized.get(b1);
                        q = binarized.get(b2);
                        break out;
                    }
                }
            }

            if (feasible) {
                min = t;
            } else {
                max = t - 1;
            }
        }

        System.out.println((p+1) + " " + (q+1));
    }
}