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 ll = long long;

// https://atcoder.jp/contests/abc142/submissions/8554606 を参考
// 最小閉路を探せばよいので全ての頂点からdfs

vector<int> graph[1001];
vector<int> ans, cur;
bitset<1001>flag;
void dfs(int v){
    if(flag[v]) return;
    flag[v] = 1;
    cur.push_back(v);
    for (int x: graph[v]) {
        if(cur[0] == x){
            if(ans.size() == 0 || ans.size() > cur.size()) ans = cur;
        } else {
            dfs(x);
        }
    }
    cur.pop_back();
}

int main() {
    int N, M;
    cin >> N >> M;
    for (int i = 0; i < M; ++i) {
        int A, B;
        cin >> A >> B;
        graph[A].push_back(B);
    }

    for (int i = 1; i <= N; ++i) {
        flag.reset();
        dfs(i);
    }

    if(ans.size() == 0) {
        cout << -1 << "\n";
        return 0;
    }
    cout << ans.size() << "\n";
    for (int x: ans) {
        cout << x << "\n";
    }
    return 0;
}