Problem: Given is a directed graph G with N vertices and M edges.
The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.
It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.
Here the null graph is not considered as a subgraph.

Constraints

* 1 \leq N \leq 1000
* 0 \leq M \leq 2000
* 1 \leq A_i,B_i \leq N
* A_i \neq B_i
* All pairs (A_i, B_i) are distinct.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_M B_M


Output

If there is no induced subgraph of G that satisfies the condition, print `-1`. Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:


K
v_1
v_2
:
v_K


This represents the induced subgraph of G with K vertices whose vertex set is \\{v_1, v_2, \ldots, v_K\\}. (The order of v_1, v_2, \ldots, v_K does not matter.) If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Examples

Input

4 5
1 2
2 3
2 4
4 1
4 3


Output

3
1
2
4


Input

4 5
1 2
2 3
2 4
1 4
4 3


Output

-1


Input

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2


Output

4
2
3
4
5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using lint = long long int;

struct Vertex {
    vector<lint> edge;
    bool visited;
    lint prev;
};
vector<Vertex> g;

// 頂点sから始まる、最小閉路の頂点集合を返す
vector<lint> bfs(lint s) {
    for(auto& e : g)  e.visited = false, e.prev = -1;

    queue<lint> que;
    que.push(s);

    while(!que.empty()) {
        lint node = que.front(); que.pop();

        for(auto next : g[node].edge) {
            if(g[next].visited)  continue;
            g[next].visited = true;
            g[next].prev = node;
            que.push(next);
        }
    }

    if(g[s].prev == -1)  return {};

    lint node = s;
    vector<lint> res;
    do {
        res.push_back(node = g[node].prev);
    } while(node != s);

    return res;
}

int main() {
    lint n, m;
    cin >> n >> m;

    g.resize(n);
    for(lint i=0; i<m; i++) {
        lint a, b;
        cin >> a >> b;
        a--, b--;
        g[a].edge.push_back(b);
    }

    lint mn = 1e18;
    vector<lint> ans;
    for(lint s=0; s<n; s++) {
        auto cand = bfs(s);
        if(cand.size() == 0)  continue;
        if(cand.size() < mn) {
            mn = cand.size();
            ans = cand;
        }
    }

    cout << (mn==1e18 ? -1 : mn) << endl;
    for(auto& e : ans)  cout << ++e << endl;
}