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.*;
import java.io.*;

public class Main {

	public static void main(String[] args) throws IOException{
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		
		String s[] = bf.readLine().split(" ");
		int n = Integer.parseInt(s[0]), m = Integer.parseInt(s[1]);
		
		int a[][] = new int[n][m];
		
		for(int i=0;i<n;i++){
			s = bf.readLine().split(" ");
			for(int j=0;j<m;j++){
				
				a[i][j] = Integer.parseInt(s[j]);
			}
		}
		
		int l = 0, r = (int)1e9;
		int a1 = 0, a2 = 0;
		int bit[] = new int[n];
		int idx[] = new int[1<<m];
		while(l<=r){
			int t = (l+r)/2;
			
			Arrays.fill(idx, 0);
			for(int i=0;i<n;i++){
				int b = 0;
				for(int j=0;j<m;j++){
					if(a[i][j] >= t)
						b |= 1<<j;
				}
				bit[i] = b;
				idx[b] = i+1;
			}
			
			boolean ok = false;
			for(int i=0;i<n;i++){
				for(int j=0;j<(1<<m);j++){
					if((bit[i] | j) == (1<<m) - 1 && idx[j] != 0){
						ok = true;
						a1 = i+1;
						a2 = idx[j];
					}
				}
				
				if(ok)
					break;
			}
			
			if(ok)
				l = t+1;
			else
				r = t-1;
		}
		
		System.out.println(a1+" "+a2);	
	}

}