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 sys import stdin, stdout

idx1 = 0
idx2 = 0
VALD = 0

def getminmax(n, m, a):
    l = 0
    h = 1000000009

    while l < h:
        mid = (l+h+1)//2
        exists = existsequalorbig(mid, m, a)

        if exists:
            #print(mid)
            l = mid
        else:
            h = mid-1

def existsequalorbig(mid, m, a):
    global idx1
    global idx2
    global VALD

    abw = []
    hs = set()

    for i in range(len(a)):
        v = 0
        for j in range(m):
            if a[i][j] >= mid:
                v |= 1
            v <<= 1
        v >>= 1
        if v not in hs:
            hs.add(v)
            abw.append([i, v])

    for i in range(len(abw)):
        for j in range(i, len(abw)):
            if abw[i][1] | abw[j][1] == VALD:
                idx1 = abw[i][0]
                idx2 = abw[j][0]
                return True

    #print(abw)
    #print(VALD)
    return False


#O(logA(4m+nm))
if __name__ == '__main__':
    nm = list(map(int, stdin.readline().split()))
    n = nm[0]
    m = nm[1]

    VALD = int(pow(2, m) - 1)

    a = []
    for i in range(n):
        a.append(list(map(int, stdin.readline().split())))

    getminmax(n, m, a)

    #existsequalorbig(1, m, a)

    stdout.write(str(idx1+1) + ' ' + str(idx2+1))