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 PYTHON):

Solution:
n,m=[int(i) for i in raw_input().split()]
a=[[int(i) for i in raw_input().split()] for j in range(n)]
st=0
end=10**9
o=[]
while st<=end:
    mid=(st+end)/2
    b=[]
    maxx=0
    dp={}
    for i in range(1,2**m):
        dp[i]=-1
    for i in range(n):
        val=0
        for j in range(m):
            if a[i][j]>=mid:
                val+=2**j
        b.append(val)
        dp[val]=i
    pos=-1
    pos2=-1
    for i in dp.keys():
        pos1=dp[i]
        if pos1==-1:
            continue
        val=(2**m-1)^i
        for j in dp:
            if val&j==val and dp[j]!=-1:
                pos2=dp[j]
        if pos2!=-1:
            st=mid+1
            o=[pos1+1,pos2+1]
            break
    else:
        end=mid-1
print o[0],o[1]