#include <bits/stdc++.h>

using namespace std;
const int N = 40;

int n;
int a[N][N];
int b[N][N];
int dx[] = {0, 0, -1, 1};
int dy[] = {-1, 1, 0, 0};
struct coord {
    int x, y;
};

//广度优先搜索
void bfs(int x, int y) {
    queue<coord> q;
    q.push({x, y});
    while (!q.empty()) {
        coord p = q.front();
        q.pop();
        b[p.x][p.y] = -1;

        for (int i = 0; i < 4; i++) {
            int x1 = p.x + dx[i], y1 = p.y + dy[i];
            if (b[x1][y1] >= 0 && x1 >= 1 && x1 <= n && y1 >= 1 && y1 <= n && a[x1][y1] == 0) {
                q.push({x1, y1});
                b[x1][y1] = -1;
            }
        }
    }
}

int main() {
    //读入地图
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            cin >> a[i][j];
            b[i][j] = a[i][j];
        }

    //思路：从边缘（四条外边），找0，然后BFS找出所有区域，这些区域就是不被圈起来的范围，进行标识，再取反
    for (int i = 1; i <= n; i++)
        if (a[1][i] == 0) bfs(1, i);    //第一行

    for (int i = 1; i <= n; i++)
        if (a[i][1] == 0) bfs(i, 1);    //第一列

    for (int i = 1; i <= n; i++)
        if (a[n][i] == 0) bfs(n, i);      //最后一行

    for (int i = 1; i <= n; i++)
        if (a[i][n] == 0) bfs(i, n);      //最后一列

    //输出
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (b[i][j] == -1) cout << 0 << " ";
            else if (b[i][j] == 0) cout << 2 << " ";
            else cout << b[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}