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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (int)1e9 + 7;

int main() {
    int n, m; cin >> n >> m;
    vector<vector<int>> g(n);
    while (m--) {
        int a, b; cin >> a >> b; a--, b--;
        g[a].emplace_back(b);
    }
    int mn = INF;
    vector<int> res;
    for (int r = 0; r < n; r++) {
        vector<int> d(n, -1), prv(n, -1);
        queue<int> que;
        que.emplace(r); d[r] = 0;
        while (!que.empty()) {
            int u = que.front(); que.pop();
            for (int v: g[u]) {
                if (d[v] == -1) {
                    d[v] = d[u] + 1;
                    prv[v] = u;
                    que.emplace(v);
                } else if (v == r && chmin(mn, d[u] + 1)) {
                    res.clear();
                    int cur = u;
                    while (~cur) res.emplace_back(cur), cur = prv[cur];
                }
            }
        }
    }
    if (res.empty()) {
        cout << -1 << endl;
    } else {
        cout << res.size() << endl;
        for (int v: res) cout << v + 1 << endl;
    }
    return 0;
}