#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<int> G[110], d(110);

int main() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        int u, k; cin >> u >> k;
        for (int j = 0; j < k; j++) {
            int v; cin >> v;
            G[u-1].push_back(v-1);
        }
    }

    for (int i = 0; i < n; i++) d[i] = -1;

    queue<int> que;
    que.push(0);
    d[0] = 0;
    while (!que.empty()) {
        int u = que.front(); que.pop();
        for (int v: G[u]) {
            if (d[v] == -1) {
                que.push(v);
                d[v] = d[u] + 1;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        cout << i+1 << " " << d[i] << endl;
    }
}

### Prompt

Please create a solution in CPP to the following problem:

Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.

Constraints

* $1 \leq n \leq 100$

Input

In the first line, an integer $n$ denoting the number of vertices, is given. In the next $n$ lines, adjacent lists of vertex $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree.$v_i$ are IDs of vertices adjacent to $u$.

Output

For each vertex $u$, print $id$ and $d$ in a line. $id$ is ID of vertex $u$ and $d$ is the distance from vertex $1$ to vertex $u$. If there are no path from vertex $1$ to vertex $u$, print -1 as the shortest distance. Print in order of IDs.

Example

Input

4
1 2 2 4
2 1 4
3 0
4 1 3


Output

1 0
2 1
3 2
4 1

### Response

```cpp
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<int> G[110], d(110);

int main() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        int u, k; cin >> u >> k;
        for (int j = 0; j < k; j++) {
            int v; cin >> v;
            G[u-1].push_back(v-1);
        }
    }

    for (int i = 0; i < n; i++) d[i] = -1;

    queue<int> que;
    que.push(0);
    d[0] = 0;
    while (!que.empty()) {
        int u = que.front(); que.pop();
        for (int v: G[u]) {
            if (d[v] == -1) {
                que.push(v);
                d[v] = d[u] + 1;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        cout << i+1 << " " << d[i] << endl;
    }
}

```