#include <bits/stdc++.h>
using namespace std;

constexpr int INF = numeric_limits<int>::max() / 4;

#define REP(i, stop) FOR(i, 0, stop)
#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)

signed main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<string> S(N);
	REP(i, N) cin >> S[i];
	vector<vector<int>> d(N, vector<int>(N));
	REP(i, N) REP(j, N) {
		d[i][j] = i == j ? 0 : S[i][j] == '1' ? 1 : INF;
	}
	REP(k, N) REP(i, N) REP(j, N) {
		int *ij = &d[i][j];
		int *ik = &d[i][k];
		int *kj = &d[k][j];
		if (*ik == INF) continue;
		if (*kj == INF) continue;
		if (*ij == INF) {
			*ij = *ik + *kj;
		} else {
			if ((*ij + *ik + *kj) & 1 == 1) {
				cout << -1 << endl;
				return 0;
			}
			if (*ij > *ik + *kj) {
				*ij = *ik + *kj;
			}
		}
	}
	int ans = 0;
	REP(i, N) REP(j, N) {
		if (d[i][j] != INF)
			ans = max(ans, d[i][j]);
	}
	cout << ans + 1 << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.

Input

Input is given from Standard Input in the following format:


N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}


Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Examples

Input

2
01
10


Output

2


Input

3
011
101
110


Output

-1


Input

6
010110
101001
010100
101000
100000
010000


Output

4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

constexpr int INF = numeric_limits<int>::max() / 4;

#define REP(i, stop) FOR(i, 0, stop)
#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)

signed main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<string> S(N);
	REP(i, N) cin >> S[i];
	vector<vector<int>> d(N, vector<int>(N));
	REP(i, N) REP(j, N) {
		d[i][j] = i == j ? 0 : S[i][j] == '1' ? 1 : INF;
	}
	REP(k, N) REP(i, N) REP(j, N) {
		int *ij = &d[i][j];
		int *ik = &d[i][k];
		int *kj = &d[k][j];
		if (*ik == INF) continue;
		if (*kj == INF) continue;
		if (*ij == INF) {
			*ij = *ik + *kj;
		} else {
			if ((*ij + *ik + *kj) & 1 == 1) {
				cout << -1 << endl;
				return 0;
			}
			if (*ij > *ik + *kj) {
				*ij = *ik + *kj;
			}
		}
	}
	int ans = 0;
	REP(i, N) REP(j, N) {
		if (d[i][j] != INF)
			ans = max(ans, d[i][j]);
	}
	cout << ans + 1 << endl;
}

```