#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
using namespace std;
using ll = long long;
using ld = long double;
const int N = 51;
int a[N][N], b[N][N];
int dp[N][N][N][N], dp1[N][N][N][N];
int kek;
void print(int n) {
  cout << "!" << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) {
        cout << a[i][j] - 1;
      } else {
        cout << ((a[i][j] - 1) ^ kek);
      }
    }
    cout << endl;
  }
  exit(0);
}
int ask(int x1, int y1, int x2, int y2) {
  cout << "? ";
  cout << x1 << ' ' << y1 << ' ';
  cout << x2 << ' ' << y2 << endl;
  int r;
  cin >> r;
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  a[1][1] = 2;
  a[n][n] = 1;
  a[1][2] = 1;
  a[2][3] = (ask(1, 2, 2, 3) ? 1 : 2);
  a[2][1] = (ask(2, 1, 2, 3) ? a[2][3] : 3 - a[2][3]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j]) {
        continue;
      }
      if (j >= 2 && a[i][j - 2]) {
        a[i][j] = (ask(i, j - 2, i, j) ? a[i][j - 2] : 3 - a[i][j - 2]);
        continue;
      }
      if (i >= 2 && a[i - 2][j]) {
        a[i][j] = (ask(i - 2, j, i, j) ? a[i - 2][j] : 3 - a[i - 2][j]);
        continue;
      }
      if (a[i - 1][j - 1]) {
        a[i][j] =
            (ask(i - 1, j - 1, i, j) ? a[i - 1][j - 1] : 3 - a[i - 1][j - 1]);
        continue;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((i + j) & 1) {
        b[i][j] = 3 - a[i][j];
      } else {
        b[i][j] = a[i][j];
      }
    }
  }
  vector<int> dx = {0, 1};
  vector<int> dy = {1, 0};
  for (int q = 0; q <= n - 1; q++) {
    for (int w = 0; w <= n - 1; w++) {
      for (int i = 1; i <= n - q; i++) {
        for (int j = 1; j <= n - w; j++) {
          int x = i + q;
          int y = j + w;
          if (x == i && y == j) {
            dp[i][j][x][y] = dp1[i][j][x][y] = 1;
            continue;
          }
          if ((x == i + 1 && y == j) || (x == i && y == j + 1)) {
            dp[i][j][x][y] = (a[i][j] == a[x][y]);
            dp1[i][j][x][y] = (b[i][j] == b[x][y]);
            continue;
          }
          for (int t1 = 0; t1 < 2; t1++) {
            for (int t2 = 0; t2 < 2; t2++) {
              dp[i][j][x][y] |=
                  (a[i][j] == a[x][y]) &&
                  dp[i + dx[t1]][j + dy[t1]][x - dx[t2]][y - dy[t2]];
              dp1[i][j][x][y] |=
                  (b[i][j] == b[x][y]) &&
                  dp1[i + dx[t1]][j + dy[t1]][x - dx[t2]][y - dy[t2]];
            }
          }
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int x = i; x <= n; x++) {
        for (int y = j; y <= n; y++) {
          if ((x - i + y - j + 1) & 1 || x - i + y - j <= 1) {
            continue;
          }
          if (dp[i][j][x][y]) {
            kek = 1 - ask(i, j, x, y);
            print(n);
          }
          if (dp1[i][j][x][y]) {
            kek = ask(i, j, x, y);
            print(n);
          }
        }
      }
    }
  }
  return 0;
}
