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.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		int m=s.nextInt();
		
		int[][] arr=new int[n][m];
		
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				arr[i][j]=s.nextInt();
			}
		}
		
		int index1=-1;
		int index2=-1;
		
		int start=0;
		int end=1000000000;
		
		int full=(int)(Math.pow(2,m)-1);
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			pair p=check(arr,n,m,mid,full);
			
			if(p.ans)
			{
				index1=p.index1;
				index2=p.index2;
				
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
		}
		
		System.out.println((index1+1)+" "+(index2+1));
		
	}
	
	public static pair check(int[][] arr,int n,int m,int mid,int full)
	{
		int[] count=new int[260];
		int[] index=new int[260];
		
		for(int i=0;i<n;i++)
		{
			int mask=0;
			
			for(int j=0;j<m;j++)
			{
				if(arr[i][j]>=mid)
				{
					mask=((1<<j)|mask);
				}
			}
			
			count[mask]=1;
			index[mask]=i;
		}
		
		for(int i=0;i<260;i++)
		{
			if(count[i]>0)
			{
				for(int j=i;j<260;j++)
				{
					if(count[j]>0)
					{
						if((i|j)==full)
						{
							pair p=new pair(true,index[i],index[j]);
							return p;
						}
					}
				}
			}
			
		}
		
		pair p=new pair(false,-1,-1);
		return p;	
	}
}

class pair
{
	boolean ans;
	int index1;
	int index2;
	
	public pair(boolean ans,int index1,int index2)
	{
		this.ans=ans;
		this.index1=index1;
		this.index2=index2;
	}
}