#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector> 
using namespace std;
vector<int> result;

int _5to10(const vector<int>& x){
	int num = 0;
	for(int i = x.size() - 1; i >= 0; i--){
		num = num * 5 + x[i];
	}
	return num;
}

//T取bool或者int[0, 1]，这里考虑写一个完整的4叉树的类 
template <class T>
struct Node{
	T element;
	bool isLeaf;
	Node<T>* NW;//1
	Node<T>* NE;//2
	Node<T>* SW;//3
	Node<T>* SE;//4
	vector<int> path;
	
	Node(const T& _element) : isLeaf(true), element(_element), NW(nullptr), NE(nullptr), SW(nullptr), SE(nullptr) {} 
};

//这里不需要Tree，可以直接用递归做，更方便  
//存对角线的(i, j) 
template <class T>
void dfs(vector<vector<int>>& mat, Node<T>*& node, vector<int>& path, int li, int lj, int ri, int rj) {
    int val = mat[li][lj];
    bool same = true;
    for (int i = li; i <= ri && same; i++) {
        for (int j = lj; j <= rj && same; j++) {
            if (mat[i][j] != val) same = false;
        }
    }

    if (same) {  // 叶子
        node = new Node<T>(val);
        node->path = path;
        
        //转换10进制 
        result.push_back(_5to10(node->path));
        return;
    }

    node = new Node<T>(-1); // 内部节点
    node->isLeaf = false;

    int midRow = (li + ri) / 2;
    int midCol = (lj + rj) / 2;

    // NW=1
    path.push_back(1);
    dfs(mat, node->NW, path, li, lj, midRow, midCol);
    path.pop_back();

    // NE=2
    path.push_back(2);
    dfs(mat, node->NE, path, li, midCol+1, midRow, rj);
    path.pop_back();

    // SW=3
    path.push_back(3);
    dfs(mat, node->SW, path, midRow+1, lj, ri, midCol);
    path.pop_back();

    // SE=4
    path.push_back(4);
    dfs(mat, node->SE, path, midRow+1, midCol+1, ri, rj);
    path.pop_back();
    
}

int main(){
    int n, m;
    cin >> n >> m;
    vector<vector<int>> matrix(n, vector<int>(m));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            cin >> matrix[i][j];

    Node<int>* root = nullptr;
    vector<int> path;
    dfs(matrix, root, path, 0, 0, n-1, m-1);

	sort(result.begin(), result.end());
	for (int v : result) cout << v << " ";
    // TODO: 遍历叶子节点，把 path 转换为十进制数输出
    return 0;
}
/*
8 8
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
0 0 0 1 1 1 1 1
0 0 1 1 1 1 1 1
0 0 1 1 1 1 0 0
0 0 1 1 1 0 0 0
1 7 8 9 12 14 17 18 22 23 24 38 44 63 69 88 94 113 119

*/