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>
using namespace std;
const long long INF = 1e18 + 1e18;
const long long MOD = 1000000007;
const long long MAX = 4000001;
const long double eps = 1E-14;
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long min(long long a, long long b) {
  if (a > b) {
    return b;
  }
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  if (a < b) {
    return gcd(b, a);
  }
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
struct edge {
  long long ind;
  long long fr;
  long long to;
  long long d;
};
class mint {
  long long x;

 public:
  mint(long long x = 0) : x((x % MOD + MOD) % MOD) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint& a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator-=(const mint& a) {
    if ((x += MOD - a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator*=(const mint& a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint operator+(const mint& a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint& a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint& a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(long long t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(MOD - 2); }
  mint& operator/=(const mint& a) { return (*this) *= a.inv(); }
  mint operator/(const mint& a) const {
    mint res(*this);
    return res /= a;
  }
  friend ostream& operator<<(ostream& os, const mint& m) {
    os << m.x;
    return os;
  }
};
mint pw(mint a, long long b) {
  if (b == 0) {
    return 1;
  }
  mint ret = pw(a, b >> 1);
  ret *= ret;
  if (b & 1) {
    ret *= a;
  }
  return ret;
}
int main() {
  long long N, M;
  cin >> N >> M;
  vector<vector<long long>> A(N + 1, vector<long long>(M + 1, 0));
  for (long long i = 1; i <= N; i++)
    for (long long j = 1; j <= M; j++) cin >> A[i][j];
  vector<long long> R(N + 1, 0), C(M + 1, 0);
  for (long long i = 1; i <= N; i++)
    for (long long j = 1; j <= M; j++) R[i] = max(R[i], A[i][j]);
  for (long long i = 1; i <= N; i++)
    for (long long j = 1; j <= M; j++) C[j] = max(C[j], A[i][j]);
  long long K = N * M;
  vector<long long> isr(K + 1, 0), isc(K + 1, 0);
  for (long long i = 1; i <= N; i++) isr[R[i]] = 1;
  for (long long j = 1; j <= M; j++) isc[C[j]] = 1;
  vector<vector<long long>> ans(N + 2, vector<long long>(M + 2, 0));
  long long posx = 0;
  long long posy = 0;
  queue<long long> qx, qy;
  vector<long long> dx = {1, 0, -1, 0};
  vector<long long> dy = {0, 1, 0, -1};
  for (long long k = K; k >= 1; k--) {
    long long adx, ady;
    if (isr[k] && isc[k]) {
      posx++;
      posy++;
      adx = posx;
      ady = posy;
    } else if (isr[k]) {
      posx++;
      adx = posx;
      ady = posy;
    } else if (isc[k]) {
      posy++;
      adx = posx;
      ady = posy;
    } else {
      adx = qx.front();
      qx.pop();
      ady = qy.front();
      qy.pop();
    }
    ans[adx][ady] = k;
    for (long long j = 0; j < 4; j++) {
      long long xx = adx + dx[j];
      long long yy = ady + dy[j];
      if (xx <= 0 || xx > N || yy <= 0 || yy > M) {
        continue;
      }
      if (ans[xx][yy] != 0) {
        continue;
      }
      long long count = 0;
      for (long long l = 0; l < 4; l++) {
        if (ans[xx + dx[l]][yy + dy[l]] != 0) {
          count++;
        }
      }
      if (count == 2) {
        qx.push(xx);
        qy.push(yy);
      }
    }
  }
  for (long long i = 1; i <= N; i++) {
    for (long long j = 1; j <= M; j++) cout << ans[i][j] << " ";
    cout << endl;
  }
}