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:

def main():
  N, M = map(int, input().split())
  L = [tuple(map(int, input().split())) for _ in range(N)]
  maxi = max(max(t) for t in L)+1
  mini, res = max((min(t), i) for i, t in enumerate(L))
  res = res, res
  BITMASK = (1 << M)
  while True:
    mid = (maxi+mini)//2
    #print(f'{mini} {mid} {maxi}')
    if mid == mini:
      break
    masks = [None]*BITMASK
    for i, t in enumerate(L):
      tmask = 0
      for v in t:
        tmask *= 2
        if v >= mid:
          tmask += 1
      if masks[tmask] is not None:
        continue
      masks[tmask] = i
      for k in range(BITMASK):
        if masks[k] is not None and k | tmask == BITMASK-1:
          res = masks[k], i
          mini = mid = min(max(a, b) for a, b in zip(L[res[0]], L[res[1]]))
          break
      else:
        continue
      break
    else:
      maxi = mid
    #print(masks)
  print(res[0]+1, res[1]+1)


main()