#include<bits/stdc++.h>
using namespace std;
int n;
int ask(vector<pair<int,int> > vec)
{
bool out[n][n];
memset(out, false, sizeof(out));
for(auto [x, y] : vec) out[x][y] = 1;
cout << "?" << endl;
for(int i = 0; i < n; i ++) {
for(int j = 0; j < n; j++)
cout << out[i][j];
cout << endl;
}
int res;
cin >> res;
return res;
}
void say(vector<pair<int,int> > vec)
{
bool out[n][n];
memset(out, false, sizeof(out));
for(auto [x, y] : vec) out[x][y] = 1;
cout << "!" << endl;
for(int i = 0; i < n; i ++) {
for(int j = 0; j < n; j++)
cout << out[i][j];
cout << endl;
}
exit(0);
}
void solve(int idx, int res)
{
for(int i = idx; i < n; i ++)
{
vector<pair<int,int> > on;
for(int j = 0; j < n; j ++)
on.push_back({i, j});
if(ask(on) == n * n)
say(on);
}
vector<pair<int,int> > sol;
if(res == n)
{
sol.push_back({idx, 0});
for(int i = 0; i < idx; i ++)
{
if(ask({{i, 0}, {idx, 0}}) == 2 * n)
sol.push_back({i, 0});
else
sol.push_back({i, n - 1});
}
for(int i = idx + 1; i < n; i ++)
{
if(ask({{idx, 0}, {i, 0}}) == 2 * n)
{
sol.push_back({i, 0});
continue;
}
int l = 0, r = n;
while(r - l > 1)
{
int mid = (l + r) / 2;
vector<pair<int,int> > on = {};
for(int j = 0; j < mid; j ++)
on.push_back({i, j});
if(ask(on) == mid * n)
l = mid;
else
r = mid;
}
sol.push_back({i, l});
}
say(sol);
}
for(int i = 0; i < idx; i ++)
{
if(ask({{idx, 0}, {i, 0}}) == n)
sol.push_back({i, n - 1});
else
sol.push_back({i, 0});
}
for(int i = idx; i < n; i ++)
{
int l = 0, r = n;
while(r - l > 1)
{
int mid = (l + r) / 2;
vector<pair<int,int> > on = {{idx, 0}};
for(int j = 0; j < mid; j++)
on.push_back({i, j});
if(ask(on) == mid * n)
l = mid;
else
r = mid;
}
sol.push_back({i, l});
}
say(sol);
}
int main()
{
cin >> n;
for(int i = 0; i < n; i ++)
{
int res = ask({{i, 0}, {i, n - 1}});
if(res != 2 * n - 1)
solve(i, res);
}
vector<pair<int,int> > sol = {{0, 0}}, sol2 = {{0, n-1}};
for(int i = 1; i < n; i ++) // if the first is <->
{
int res1 = ask({{0, 0}, {i, 0}});
int res2 = ask({{0, n -1}, {i, n - 1}});
if(res1 == 2 * n)
sol.push_back({i, 0});
else
sol.push_back({i, n-1});
if(res2 == 2 * n)
sol2.push_back({i, n - 1});
else
sol2.push_back({i, 0});
}
if(ask(sol2) == n * n)
say(sol2);
say(sol);
return 0;
}
