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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class D {
	FastScanner in;
	PrintWriter out;
	boolean systemIO = true;

	public class Pair implements Comparable<Pair> {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			if (x > o.x) {
				return 1;
			}
			if (x < o.x) {
				return -1;
			}
			if (y > o.y) {
				return 1;
			}
			if (y < o.y) {
				return -1;
			}
			return 0;
		}

	}

	public class Fenvik {
		int[] sum;

		public Fenvik(int n) {
			sum = new int[n];
		}

		public void add(int x, int d) {
			for (int i = x; i < sum.length; i = (i | (i + 1))) {
				sum[i] += d;
			}
		}

		public int sum(int r) {
			int ans = 0;
			for (int i = r; i >= 0; i = (i & (i + 1)) - 1) {
				ans += sum[i];
			}
			return ans;
		}

		public int sum(int l, int r) {
			if (l > r) {
				return 0;
			}
			return sum(r) - sum(l - 1);
		}
	}

	public long gcd(long x, long y) {
		if (y == 0) {
			return x;
		}
		if (x == 0) {
			return y;
		}
		return gcd(y, x % y);
	}

	public long[][] pow(long[][] x, long p) {
		if (p == 0) {
			long[][] ans = new long[x.length][x.length];
			for (int i = 0; i < ans.length; i++) {
				ans[i][i] = 1;
			}
			return ans;
		}
		long[][] t = pow(x, p / 2);
		t = multiply(t, t);
		if (p % 2 == 1) {
			t = multiply(t, x);
		}
		return t;
	}

	public long[][] multiply(long[][] a, long[][] b) {
		long[][] ans = new long[a.length][b[0].length];
		for (int i = 0; i < ans.length; i++) {
			for (int j = 0; j < ans[0].length; j++) {
				for (int k = 0; k < b.length; k++) {
					ans[i][j] += a[i][k] * b[k][j];
					ans[i][j] %= mod;
				}
			}
		}
		return ans;
	}

	public long pow(long x, long p) {
		if (p == 0) {
			return 1;
		}
		long t = pow(x, p / 2);
		t *= t;
		t %= mod;
		if (p % 2 == 1) {
			t *= x;
			t %= mod;
		}
		return t;
	}

	public long modInv(long x) {
		return pow(x, mod - 2);
	}

	long mod = 1000000007;

	Random random = new Random(566);

	public void solve() {
		int n = in.nextInt();
		int m = in.nextInt();
		int[][] a = new int[n][m];
		for (int i = 0; i < a.length; i++) {
			for (int j = 0; j < a[0].length; j++) {
				a[i][j] = in.nextInt();
			}
		}
		TreeSet<Integer> free = new TreeSet<>();
		for (int i = 1; i <= n * m; i++) {
			free.add(i);
		}
		TreeSet<Integer> r = new TreeSet<>();
		TreeSet<Integer> c = new TreeSet<>();
		for (int i = 0; i < a.length; i++) {
			int max = 0;
			for (int j = 0; j < a[0].length; j++) {
				max = Math.max(max, a[i][j]);
			}
			r.add(max);
			free.remove(max);
		}
		for (int i = 0; i < a[0].length; i++) {
			int max = 0;
			for (int j = 0; j < a.length; j++) {
				max = Math.max(max, a[j][i]);
			}
			c.add(max);
			free.remove(max);
		}
		int[][] b = new int[n][m];
		int x = -1;
		int y = -1;
		Stack<Pair> st = new Stack<>();
		while (!r.isEmpty() && !c.isEmpty()) {
			int max = Math.max(r.last(), c.last());
			if (max == r.last()) {
				x++;
				r.pollLast();
			}
			if (max == c.last()) {
				y++;
				c.pollLast();
			}
			b[x][y] = max;
			st.add(new Pair(x, y));
		}
		while (!r.isEmpty()) {
			x++;
			b[x][y] = r.pollLast();
			st.add(new Pair(x, y));
		}
		while (!c.isEmpty()) {
			y++;
			b[x][y] = c.pollLast();
			st.add(new Pair(x, y));
		}
		while (!st.isEmpty()) {
			Pair p = st.pop();
			for (int j = 0; b[p.x][j] == 0; j++) {
				b[p.x][j] = free.pollFirst();
			}
			for (int i = 0; b[i][p.y] == 0; i++) {
				b[i][p.y] = free.pollFirst();
			}
		}
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < b[0].length; j++) {
				out.print(b[i][j] + " ");
			}
			out.println();
		}
	}

	public void run() {
		try {
			if (systemIO) {
				in = new FastScanner(System.in);
				out = new PrintWriter(System.out);
			} else {
				in = new FastScanner(new File("input.txt"));
				out = new PrintWriter(new File("output.txt"));
			}
			solve();

			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}

		String nextLine() {
			try {
				return br.readLine();
			} catch (IOException e) {
				return null;
			}
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

	}

	// AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
	public static void main(String[] arg) {
		long time = System.currentTimeMillis();
		new D().run();
		System.err.println(System.currentTimeMillis() - time);
	}
}