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;
struct data {
  int x, y;
} d[int(300 + 5) * 4];
int N, M, a[int(300 + 5)][int(300 + 5)], b[int(300 + 5)][int(300 + 5)], x, y,
    num;
int R[int(300 + 5)], C[int(300 + 5)], cnt[int(300 + 5) * int(300 + 5)];
inline void read(int &x) {
  x = 0;
  bool f = 0;
  char ch = getchar();
  for (; !(ch >= '0' && ch <= '9'); ch = getchar()) f |= ch == '-';
  for (; (ch >= '0' && ch <= '9'); ch = getchar())
    x = (x << 3) + (x << 1) + (ch ^ 48);
  x = f ? -x : x;
}
void sol(int i, int j) {
  for (int k = j; k >= 1; --k) {
    if (b[i][k]) continue;
    while (cnt[num] == 0 && num > 0) --num;
    b[i][k] = num;
    cnt[num] = 0;
    --num;
  }
  for (int k = i; k >= 1; --k) {
    if (b[k][j]) continue;
    while (cnt[num] == 0 && num > 0) --num;
    b[k][j] = num;
    cnt[num] = 0;
    --num;
  }
}
int main() {
  read(N);
  read(M);
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= M; ++j)
      read(a[i][j]), R[i] = ((R[i]) > (a[i][j]) ? (R[i]) : (a[i][j])),
                     C[j] = ((C[j]) > (a[i][j]) ? (C[j]) : (a[i][j])),
                     cnt[(i - 1) * M + j] = 1;
  }
  sort(R + 1, R + 1 + N);
  sort(C + 1, C + 1 + M);
  for (int i = 1; i <= N / 2; ++i) swap(R[i], R[N - i + 1]);
  for (int i = 1; i <= M / 2; ++i) swap(C[i], C[M - i + 1]);
  for (int i = 1; i <= N; ++i) cnt[R[i]] = 0;
  for (int i = 1; i <= M; ++i) cnt[C[i]] = 0;
  x = 1;
  y = 1;
  num = 0;
  for (; x <= N || y <= M;) {
    if (x != N + 1 || y != M + 1) {
      if (R[x] == C[y]) {
        b[x][y] = R[x];
        cnt[R[x]] = 0;
        d[++num] = (data){x, y};
        ++x;
        ++y;
      } else if (R[x] < C[y]) {
        b[x - 1][y] = C[y];
        d[++num] = (data){x - 1, y};
        cnt[C[y]] = 0;
        ++y;
      } else {
        b[x][y - 1] = R[x];
        d[++num] = (data){x, y - 1};
        cnt[R[x]] = 0;
        ++x;
      }
    }
  }
  num = N * M;
  while (cnt[num] == 0 && num > 0) --num;
  for (int i = 1, j = 1, l = 1; num;) {
    if (d[l].x + 1 == d[l + 1].x && d[l].y + 1 == d[l + 1].y) {
      ++i;
      ++j;
      sol(i, j);
      ++l;
      continue;
    } else {
      if (d[l].x + 1 == d[l + 1].x && d[l].y == d[l + 1].y) {
        while (d[l].x + 1 == d[l + 1].x && d[l].y == d[l + 1].y) ++l;
        int x = d[l].x, y = d[l].y;
        for (int k = i + 1; k <= x; ++k) {
          for (int u = y - 1; u >= 1; --u) {
            if (b[k][u]) continue;
            while (cnt[num] == 0 && num > 0) --num;
            b[k][u] = num;
            cnt[num] = 0;
            --num;
          }
        }
        i = x;
        j = y;
      } else {
        while (d[l].x == d[l + 1].x && d[l].y + 1 == d[l + 1].y) ++l;
        int x = d[l].x, y = d[l].y;
        for (int k = j + 1; k <= y; ++k) {
          for (int u = x - 1; u >= 1; --u) {
            if (b[u][k]) continue;
            while (cnt[num] == 0 && num > 0) --num;
            b[u][k] = num;
            cnt[num] = 0;
            --num;
          }
        }
        i = x;
        j = y;
      }
    }
  }
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= M; ++j) printf("%d%c", b[i][j], j == M ? '\n' : ' ');
  }
  return 0;
}