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.io.*;
import java.util.*;

public class d {
	public static void main(String[] args) { new d(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int n, m, oo = (int)1e9;
	int[][] a;

	d() {
		n = in.nextInt();
		m = in.nextInt();

		a = new int[m][n];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[j][i] = in.nextInt();
			
		int[][] minmask = new int[1 << m][n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < (1 << m); j++) {
				int min = oo;
				for (int k = 0; k < m; k++)
					if ((j & (1 << k)) > 0)
						min = min(min, a[k][i]);
				minmask[j][i] = min;
			}
		}
		
		int[] maxmask = new int[1 << m];
		for (int i = 0; i < (1 << m); i++) {
			int max = -1, k = -1;
			for (int j = 0; j < n; j++) {
				if (minmask[i][j] > max) {
					max = minmask[i][j];
					k = j;
				}
			}
			maxmask[i] = k;
		}

		int ans = -1, x = -1, y = -1;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < (1 << m); j++) {
				int u = ((1 << m) - 1) ^ j;
				int idx = maxmask[u], max = oo;
				for (int k = 0; k < m; k++) max = min(max, max(a[k][i], a[k][idx]));
				if (max > ans) {
					ans = max;
					x = i;
					y = idx;
				}
			}
		}

		x++; y++;
		out.println(x + " " + y);
		out.close();
	}

	
	int abs(int x) { if (x < 0) return -x; return x; }
	long abs(long x) { if (x < 0) return -x; return x; }
	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	
	int gcd(int a, int b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long gcd(long a, long b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long lcm(int a, int b) { return (((long) a) * b) / gcd(a, b); }
	long lcm(long a, long b) { return (a * b) / gcd(a, b); }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}