Problem: Koa the Koala has a matrix A of n rows and m columns. Elements of this matrix are distinct integers from 1 to n ⋅ m (each number from 1 to n ⋅ m appears exactly once in the matrix).

For any matrix M of n rows and m columns let's define the following:

  * The i-th row of M is defined as R_i(M) = [ M_{i1}, M_{i2}, …, M_{im} ] for all i (1 ≤ i ≤ n). 
  * The j-th column of M is defined as C_j(M) = [ M_{1j}, M_{2j}, …, M_{nj} ] for all j (1 ≤ j ≤ m). 



Koa defines S(A) = (X, Y) as the spectrum of A, where X is the set of the maximum values in rows of A and Y is the set of the maximum values in columns of A.

More formally:

  * X = \{ max(R_1(A)), max(R_2(A)), …, max(R_n(A)) \} 
  * Y = \{ max(C_1(A)), max(C_2(A)), …, max(C_m(A)) \}



Koa asks you to find some matrix A' of n rows and m columns, such that each number from 1 to n ⋅ m appears exactly once in the matrix, and the following conditions hold:

  * S(A') = S(A) 
  * R_i(A') is bitonic for all i (1 ≤ i ≤ n) 
  * C_j(A') is bitonic for all j (1 ≤ j ≤ m) 

An array t (t_1, t_2, …, t_k) is called bitonic if it first increases and then decreases.

More formally: t is bitonic if there exists some position p (1 ≤ p ≤ k) such that: t_1 < t_2 < … < t_p > t_{p+1} > … > t_k.

Help Koa to find such matrix or to determine that it doesn't exist.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 250) — the number of rows and columns of A.

Each of the ollowing n lines contains m integers. The j-th integer in the i-th line denotes element A_{ij} (1 ≤ A_{ij} ≤ n ⋅ m) of matrix A. It is guaranteed that every number from 1 to n ⋅ m appears exactly once among elements of the matrix.

Output

If such matrix doesn't exist, print -1 on a single line.

Otherwise, the output must consist of n lines, each one consisting of m space separated integers — a description of A'.

The j-th number in the i-th line represents the element A'_{ij}.

Every integer from 1 to n ⋅ m should appear exactly once in A', every row and column in A' must be bitonic and S(A) = S(A') must hold.

If there are many answers print any.

Examples

Input


3 3
3 5 6
1 7 9
4 8 2


Output


9 5 1
7 8 2
3 6 4


Input


2 2
4 1
3 2


Output


4 1
3 2


Input


3 4
12 10 8 6
3 4 5 7
2 11 9 1


Output


12 8 6 1
10 11 9 2
3 4 5 7

Note

Let's analyze the first sample:

For matrix A we have:

    * Rows: 
      * R_1(A) = [3, 5, 6]; max(R_1(A)) = 6 
      * R_2(A) = [1, 7, 9]; max(R_2(A)) = 9 
      * R_3(A) = [4, 8, 2]; max(R_3(A)) = 8 

    * Columns: 
      * C_1(A) = [3, 1, 4]; max(C_1(A)) = 4 
      * C_2(A) = [5, 7, 8]; max(C_2(A)) = 8 
      * C_3(A) = [6, 9, 2]; max(C_3(A)) = 9 

  * X = \{ max(R_1(A)), max(R_2(A)), max(R_3(A)) \} = \{ 6, 9, 8 \} 
  * Y = \{ max(C_1(A)), max(C_2(A)), max(C_3(A)) \} = \{ 4, 8, 9 \} 
  * So S(A) = (X, Y) = (\{ 6, 9, 8 \}, \{ 4, 8, 9 \}) 



For matrix A' we have:

    * Rows: 
      * R_1(A') = [9, 5, 1]; max(R_1(A')) = 9 
      * R_2(A') = [7, 8, 2]; max(R_2(A')) = 8 
      * R_3(A') = [3, 6, 4]; max(R_3(A')) = 6 

    * Columns: 
      * C_1(A') = [9, 7, 3]; max(C_1(A')) = 9 
      * C_2(A') = [5, 8, 6]; max(C_2(A')) = 8 
      * C_3(A') = [1, 2, 4]; max(C_3(A')) = 4 

  * Note that each of this arrays are bitonic. 
  * X = \{ max(R_1(A')), max(R_2(A')), max(R_3(A')) \} = \{ 9, 8, 6 \} 
  * Y = \{ max(C_1(A')), max(C_2(A')), max(C_3(A')) \} = \{ 9, 8, 4 \} 
  * So S(A') = (X, Y) = (\{ 9, 8, 6 \}, \{ 9, 8, 4 \}) 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using datas = pair<ll, ll>;
using ddatas = pair<long double, long double>;
using tdata = pair<ll, datas>;
using vec = vector<ll>;
using mat = vector<vec>;
using pvec = vector<datas>;
using pmat = vector<pvec>;
const ll mod = 1000000007;
const ll inf = 1LL << 60;
const long double PI = acos(-1);
const long double eps = 1e-9;
template <class T>
inline bool chmax(T& a, T b) {
  bool x = a < b;
  if (x) a = b;
  return x;
}
template <class T>
inline bool chmin(T& a, T b) {
  bool x = a > b;
  if (x) a = b;
  return x;
}
void startupcpp() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long double distance(ddatas x, ddatas y) {
  long double a = x.first - y.first, b = x.second - y.second;
  return sqrt(a * a + b * b);
}
ll modinv(ll a) {
  ll b = mod, u = 1, v = 0, t;
  while (b) {
    t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (u + mod) % mod;
}
ll moddevide(ll a, ll b) { return (a * modinv(b)) % mod; }
vec modncrlistp, modncrlistm;
ll modncr(ll n, ll r) {
  if (n < r) return 0;
  ll i, size = modncrlistp.size();
  if (size <= n) {
    modncrlistp.resize(n + 1);
    modncrlistm.resize(n + 1);
    if (!size) {
      modncrlistp[0] = modncrlistm[0] = 1;
      size++;
    }
    for (i = size; i < (ll)n + 1; ++i) {
      modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[i] = modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
}
ll modpow(ll a, ll n) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (!b) return abs(a);
  return (a % b == 0) ? abs(b) : gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countdigits(ll n) {
  ll ans = 0;
  while (n) {
    n /= 10;
    ans++;
  }
  return ans;
}
ll sumdigits(ll n) {
  ll ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
void topcoder(vector<int>& v) {
  string s;
  while (1) {
    cin >> s;
    int i = s[0] == '{', x = 0;
    while (s[i] >= '0' && s[i] <= '9') {
      x = x * 10 + s[i] - '0';
      ++i;
    }
    v.emplace_back(x);
    if (s[i] == '}') break;
  }
}
void topcoder(string& s) {
  string t;
  cin >> t;
  int i = 1;
  while (t[i] != '"') {
    s += t[i++];
  }
}
int main() {
  int i, j, k, N, M;
  cin >> N >> M;
  vector<int> h(N), w(M, 0);
  vector<vector<int>> g(N, vector<int>(M)), ans(N, vector<int>(M, -1));
  for (i = 0; i < (ll)N; ++i) {
    for (j = 0; j < (ll)M; ++j) cin >> g[i][j];
    h[i] = *max_element((g[i]).begin(), (g[i]).end());
  }
  for (i = 0; i < (ll)N; ++i)
    for (j = 0; j < (ll)M; ++j) chmax(w[j], g[i][j]);
  sort((h).rbegin(), (h).rend());
  sort((w).rbegin(), (w).rend());
  i = j = 0;
  h.emplace_back(-1);
  w.emplace_back(-1);
  set<int> se;
  vector<int> back;
  for (k = N * M + 1, --k; k >= (ll)1; --k) {
    if (min(h[i], w[j]) != k) {
      se.insert(k);
      continue;
    } else {
      ans[i][j] = k;
      if (h[i + 1] == w[j + 1]) {
        ++i;
        ++j;
        back.emplace_back(0);
      } else if (h[i + 1] < w[j + 1]) {
        ++j;
        back.emplace_back(1);
      } else {
        ++i;
        back.emplace_back(-1);
      }
    }
  }
  reverse((back).begin(), (back).end());
  --i;
  --j;
  auto itr = se.begin();
  for (auto x : back) {
    if (x <= 0) {
      for (k = 0; k < (ll)M; ++k) {
        if (ans[i][k] < 0) {
          ans[i][k] = *itr;
          itr = se.erase(itr);
        }
      }
      --i;
    }
    if (x >= 0) {
      for (k = 0; k < (ll)N; ++k) {
        if (ans[k][j] < 0) {
          ans[k][j] = *itr;
          itr = se.erase(itr);
        }
      }
      --j;
    }
  }
  do {
    for (auto outimat : ans) do {
        bool f = 0;
        for (auto outi : outimat) {
          cout << (f ? " " : "") << outi;
          f = 1;
        }
        cout << "\n";
      } while (0);
  } while (0);
}