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;

void solve() {
    int n,m;
    cin >> n >> m;
    vector<vector<int>> g(n);
    for (int _ = 0; _ < m; _++) {
        int x,y;
        cin >> x >> y;
        x--;y--;
        g[x].emplace_back(y);
    }

    int root;

    const int INF = 0x3f3f3f3f;
    int mi_len = INF;
    vector<int> trk(n, 0);
    vector<int> pth;
    vector<int> res;
    function<void(int)> dfs = [&](int u){
        trk[u] = 1;
        pth.emplace_back(u);
        for (int v: g[u]){
            if (v == root && (int)pth.size() < mi_len) {
                mi_len = pth.size();
                res.clear();
                res = pth;
            }
            if (!trk[v]) dfs(v);
        }
        pth.pop_back();
    };
    for (int i = 0; i < n; i++) {
        fill(trk.begin(), trk.end(), 0);
        root = i; dfs(i);
    }
    if (mi_len < INF) {
        cout << mi_len << "\n";
        for (auto& x: res) {
            cout << (x+1) << "\n";
        }
    } else cout << -1;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}