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:
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n,m=map(int,input().split())
a=[list(map(int,input().split())) for i in range(n)]
l=-1;r=10**9+1
ans1,ans2=-1,-1
while r-l>1:
    x=(l+r)//2
    idx={}
    for i in range(n):
        v=0
        for j in range(m):
            if a[i][j]>=x:
                v+=1
            v<<=1
        idx[v>>1]=i
    ok=False
    idx1,idx2=0,0
    for aa,bb in idx.items():
        for cc,dd in idx.items():
            for d in range(m):
                if (aa|cc)==(2**m)-1:
                    ok=True
                    idx1=bb+1
                    idx2=dd+1
    if ok:
        l=x
        ans1=idx1
        ans2=idx2
    else:
        r=x
print(ans1,ans2)