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;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int N = 1010;
int n, m, a, b;
vector<int> adj[N], ans(N + 1);

vector<int> minCircle(int s) {
    vector<int> visited(N, 0), parent(N, 0);
    queue<int> q;
    q.push(s);
    while (!q.empty()) {
        int from = q.front();
        q.pop();
        for (int to : adj[from]) {
            if (visited[to] == 0) {
                parent[to] = from;
                visited[to] = 1;
                q.push(to);
            }
        }
        if (visited[s]) break;
    }
    int p = parent[s];
    vector<int> ret;
    ret.push_back(s);
    while (p && p != s) {
        ret.push_back(p);
        p = parent[p];
    }
    return ret;
}

int main() {
    cin >> n >> m;
    while (m--) {
        cin >> a >> b;
        adj[a].push_back(b);
    }
    for (int i = 1; i <= n; i++) {
        vector<int> t = minCircle(i);
        if (t.size() > 1 && t.size() < ans.size()) ans = t;
    }
    if (ans.size() == N + 1)
        puts("-1");
    else {
        cout << ans.size() << endl;
        for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << endl;
    }
}