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.awt.Point;
import java.util.*;
public class d {
static Scanner in=new Scanner(System.in);
	public static void main(String[] args) {
		int n=in.nextInt(),m=in.nextInt();
		int[][]d=new int[n][m];
		for(int i=0;i<n;i++)for(int j=0;j<m;j++)d[i][j]=in.nextInt();
		int[][] mins=new int[1<<d[0].length][2];//0=min, 1=index
		for(int i=0;i<1<<d[0].length;i++)for(int j=0;j<d.length;j++) {
			int min=Integer.MAX_VALUE;
			for(int k=0;k<d[0].length;k++)if(((i>>k)&1)>0)min=Math.min(min, d[j][k]);
			if (min>mins[i][0]) {
				mins[i][0]=min;
				mins[i][1]=j;
			}
		}
		int max=0;
		Point maxi=new Point(0,0);
		for(int i=0;i<1<<d[0].length;i++) {//mask 1
			int other=((1<<d[0].length)-1) ^ i;//mask 2
			int value=Math.max(max, Math.min(mins[i][0], mins[other][0]));
			if (value >max) {
				max=value;
				maxi=new Point(mins[i][1],mins[other][1]);
			}
		}
		System.out.printf("%d %d",maxi.x+1,maxi.y+1);
	}
}