#include<iostream>
using namespace std;
int a[9][9], b[9][9];
//输出
void output(int a[][9]) {
	for (int i = 0; i < 9; i++) {
		int j = 0;
		for (; j < 8; j++) {
			cout << a[i][j] << " ";
		}
		cout << a[i][j] << endl;
	}
}
//一天繁殖，题目给的最大繁殖天数为4；由内往外繁殖，子细菌不会出现在矩阵外的，故非0侧边一定能到达的，不再用if判断
void repro(int old[][9], int gener[][9]) {
	fill(gener[0], gener[0] + 81, 0);
	for (int i = 0; i < 9; i++) {
		for (int j = 0; j < 9; j++) {
			if (old[i][j] != 0) {
				//左侧
				gener[i][j - 1] += old[i][j];
				//左上
				gener[i - 1][j - 1] += old[i][j];

				//左下
				gener[i + 1][j - 1] += old[i][j];
				//右侧
				gener[i][j + 1] += old[i][j];
				//右上
				gener[i - 1][j + 1] += old[i][j];
				//右下
				gener[i + 1][j + 1] += old[i][j];
				//上侧
				gener[i - 1][j] += old[i][j];
				//下侧
				gener[i + 1][j] += old[i][j];
				//务必用加等号，不能直接用等。因为同坐标的新位置处完全有可能已经有了前面繁殖来的子细菌。该位置细菌数为原来2倍
				gener[i][j] += 2 * old[i][j];
			}
		}
	}
}

int main()
{
	int d;
	//freopen("in.txt", "r", stdin);
	cin >> a[4][4] >> d;
	for (int t = 1; t <= d; t++) {
		//旧新矩阵
		if (t % 2 == 1) {
			repro(a, b);
		}
		else {
			repro(b, a);
		}
	}
	//选择输出矩阵
	if (d % 2 == 1) {
		output(b);
	}
	else {
		output(a);
	}

	return 0;
}