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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF1288D extends PrintWriter {
	CF1288D() { super(System.out, true); }
	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		int k, l; byte[] bb = new byte[1 << 15];
		byte getc() {
			if (k >= l) {
				k = 0;
				try { l = in.read(bb); } catch (IOException e) { l = 0; }
				if (l <= 0) return -1;
			}
			return bb[k++];
		}
		int nextInt() {
			byte c = 0; while (c <= 32) c = getc();
			int a = 0;
			while (c > 32) { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1288D o = new CF1288D(); o.main(); o.flush();
	}

	static final int INF = 0x3f3f3f3f;
	int min_(int[] aa, int n, int b) {
		int a = INF;
		for (int i = 0; i < n; i++)
			if ((b & 1 << i) != 0)
				a = Math.min(a, aa[i]);
		return a;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[][] aa = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				aa[i][j] = sc.nextInt();
		int a_ = -1, i_ = -1, j_ = -1;
		for (int p = 0; p < 1 << m; p++) {
			int ip = -1, ap = -1;
			for (int i = 0; i < n; i++) {
				int a = min_(aa[i], m, p);
				if (ap < a) {
					ap = a;
					ip = i;
				}
			}
			int q = p ^ (1 << m) - 1;
			int iq = -1, aq = -1;
			for (int i = 0; i < n; i++) {
				int a = min_(aa[i], m, q);
				if (aq < a) {
					aq = a;
					iq = i;
				}
			}
			int a = Math.min(ap, aq);
			if (a_ < a) {
				a_ = a; i_ = ip; j_ = iq;
			}
		}
		println((i_ + 1) + " " + (j_ + 1));
	}
}