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 PYTHON3):

Solution:
from collections import defaultdict
from sys import stdin
input = stdin.readline
def check(mid, m):
    d = defaultdict(int)
    for idx, i in enumerate(a):
        string = ''
        for j in i:
            if j >= mid:
                string+='1'
            else:
                string+='0'
        d[int(string, 2)] = idx
    for i in d.keys():
        for j in d.keys():
            if i|j == 2**m - 1:
                 return [d[i], d[j]]
    return []
def binarySearch(lo, hi, m):
    ans = []
    while lo < hi:
        mid = lo + (hi-lo+1)//2
        x = check(mid, m)
        if x:
            lo = mid
            ans = [x[0]+1, x[1]+1]
        else:
            hi = mid-1
    return ans
n, m = map(int, input().split())
a = []
for i in range(n):
    a.append(list(map(int, input().split())))
print(*binarySearch(-1, 10**9+1, m))