#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1e7;
void fun(long long in, long long a, long long in2, long long b) {
  if (a == b) {
    cout << in + 1;
  } else if ((a == 2 && b == 3) || (a == 3 && b == 2)) {
    cout << in + 1;
  } else {
    if (b == 2 || b == 3) {
      long long x = b, y = in2;
      b = a;
      in2 = in;
      a = x;
      in = y;
    }
    if ((a == 2 && b == 1) || (a == 3 && b == 0)) {
      cout << in + 1;
    } else {
      cout << in2 + 1;
    }
  }
}
signed main() {
  cin.tie(0), ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> arr(n, vector<long long>(m));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      cin >> arr[i][j];
    }
  }
  vector<long long> ma1(n);
  vector<vector<long long>> mas(n, vector<long long>(m));
  for (long long i = 0; i < n; ++i) {
    vector<pair<long long, long long>> ne;
    for (long long j = 0; j < m; ++j) {
      ne.push_back({arr[i][j], j});
    }
    sort(ne.begin(), ne.end());
    long long pr = ne[0].first, in = 1;
    for (long long j = 0; j < m; ++j) {
      if (ne[j].first > pr) {
        in++;
        pr = ne[j].first;
      }
      mas[i][ne[j].second] = in;
    }
    ma1[i] = in;
  }
  vector<long long> ma2(m);
  vector<vector<long long>> mas2(m, vector<long long>(n));
  for (long long i = 0; i < m; ++i) {
    vector<pair<long long, long long>> ne;
    for (long long j = 0; j < n; ++j) {
      ne.push_back({arr[j][i], j});
    }
    sort(ne.begin(), ne.end());
    long long pr = ne[0].first, in = 1;
    for (long long j = 0; j < n; ++j) {
      if (ne[j].first > pr) {
        in++;
        pr = ne[j].first;
      }
      mas2[i][ne[j].second] = in;
    }
    ma2[i] = in;
  }
  vector<vector<long long>> ans(n, vector<long long>(m));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      long long nu = max(mas[i][j], mas2[j][i]);
      ans[i][j] = max(nu - mas[i][j] + ma1[i], nu - mas2[j][i] + ma2[j]);
      cout << ans[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}
