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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  if (a == 0ll) {
    return 0ll;
  }
  a %= MOD;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
void Pv(const vector<int>& V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%d", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
void Pvl(const vector<long long>& V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%lld", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
int n, m;
int a[300010][9];
unordered_set<int> st;
int t1, t2;
bool check(int x) {
  st.clear();
  int t;
  for (int i = 1; i <= n; i++) {
    t = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= x) {
        t |= (1 << (j - 1));
      }
    }
    st.insert(t);
  }
  for (auto y : st) {
    for (auto z : st) {
      if ((y | z) == (1 << m) - 1) {
        t1 = y;
        t2 = z;
        return 1;
      }
    }
  }
}
int main() {
  scanf("%d %d", &(n), &(m));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &(a[i][j]));
      ;
    }
  }
  int l = 0;
  int r = 1e9;
  int mid;
  int ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid)) {
      l = mid + 1;
      ans = mid;
    } else {
      r = mid - 1;
    }
  }
  int ans1, ans2;
  for (int i = 1; i <= n; i++) {
    int t = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= ans) {
        t |= (1 << (j - 1));
      }
    }
    if (t == t1) {
      ans1 = i;
    }
    if (t == t2) {
      ans2 = i;
    }
  }
  printf("%d %d\n", ans1, ans2);
  return 0;
}
inline void getInt(int* p) {
  char ch;
  do {
    ch = getchar();
  } while (ch == ' ' || ch == '\n');
  if (ch == '-') {
    *p = -(getchar() - '0');
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 - ch + '0';
    }
  } else {
    *p = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') {
      *p = *p * 10 + ch - '0';
    }
  }
}