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 \}) 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Solution {
    static final FS sc = new FS();
    static final PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) {
        int n = sc.nextInt(), m = sc.nextInt();
        int[][] matrix = new int[n][m];
        for (int i = 0;i < n;i++) {
            matrix[i] = sc.nextArray(m);
        }
        int[] h = new int[n * m + 1];
        int[] v = new int[n * m + 1];
        for (int i = 0; i < n; ++i) {
            int max = 0;
            for (int j = 0; j < m; ++j)
                max = Math.max(max, matrix[i][j]);
            h[max] = 1;
        }

        for (int i = 0; i < m; ++i) {
            int max = 0;
            for (int j = 0; j < n; ++j)
                max = Math.max(max, matrix[j][i]);
            v[max] = 1;
        }
        int[][] ans = new int[n][m];
        Queue<Node> queue = new LinkedList<>();
        int x = -1, y = -1;
        for (int u = n * m;u >= 1;u--) {
            x += h[u];
            y += v[u];
            if (h[u] > 0 || v[u] > 0) {
                ans[x][y] = u;
            } else {
                Node node = queue.poll();
                ans[node.x][node.y] = u;
            }
            if (h[u] > 0) {
                for (int i = y - 1;i >= 0;i--) {
                    queue.add(new Node(x, i));
                }
            }
            if (v[u] > 0) {
                for (int i = x - 1;i >= 0;i--) {
                    queue.add(new Node(i, y));
                }
            }
        }
        for (int i = 0;i < n;i++) {
            for (int j = 0;j < m;j++) {
                pw.print(ans[i][j] + " ");
            }
            pw.println();
        }
        pw.flush();
    }

    static class Node {
        int x, y;
        public Node(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
    static class FS {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while(!st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored) {}
            }
            return st.nextToken();
        }
        int[] nextArray(int n) {
            int[] a = new int[n];
            for(int i = 0;i < n;i++) {
                a[i] = nextInt();
            }
            return a;
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}