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#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;

typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;

typedef pair<int,int> ii;

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

const int MOD = 998244353;

ll read(){
    ll i;
    cin>>i;
    return i;
}

vi readvi(int n,int off=0,int shift=0){
    vi v(n+shift);
    rep(i,shift)v[i]=0;
    rep(i,n)v[i+shift]=read()+off;
    return v;
}

void YesNo(bool condition, bool do_exit=true) {
    if (condition)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    if (do_exit)
        exit(0);
}

const int INFTY = 1e6;

int main(void ) {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;

    rep(_,t) {
        int n;
        cin >> n;

        vector<string> a(n);
        vvi d(n, vi(n, INFTY));

        vvi adj(n);
        rep(i,n) {
            cin >> a[i];
            rep(j,n)
                if (a[i][j] == '1') {
                    adj[i].push_back(j);
                    d[i][j] = 1;
                }
            d[i][i] = 0;
        }

        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                }
            }
        }

        int L = 1;
        int U = 0;

        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                U = max(U, d[i][j]);

        while (L < U) {
            // rozw w [L,U]
            int D = (L+U) / 2;
            // czy da się D?

            vvi q(n, vi(n, -INFTY));
            rep(x, n)
                rep(v, n)
                    rep(y, n)
                        if (d[x][y] > D)
                            q[x][v] = max(q[x][v], d[v][y]);

            rep(u, n)
                rep (v, n)
                    if (u != v and a[u][v] == '0') {
                        bool ok = true;
                        rep (x, n)
                            if (q[x][v] > -INFTY and min(d[x][u]+q[x][v],d[x][v]+q[x][u]) > D) {
                                ok = false;
                                break;
                            }
                        if (ok)
                            goto success;
                    }

            // failed
            L = D + 1;
            continue;

            success:
            U = D;
        }

        cout << L << endl;
    }


    return 0;
}