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;

const int Maxn = 1005;

int n, m;
set <int> neigh[Maxn];
int dist[Maxn], par[Maxn];
int best = Maxn;
vector <int> bi;

void Solve(int r)
{
    fill(dist, dist + Maxn, Maxn); dist[r] = 0;
    vector <int> Q; Q.push_back(r);
    for (int i = 0; i < Q.size(); i++) {
        int v = Q[i];
        if (neigh[v].find(r) != neigh[v].end()) {
            if (dist[v] + 1 < best) {
                best = dist[v] + 1;
                bi.clear();
                bi.push_back(v);
                do {
                    v = par[v];
                    bi.push_back(v);
                } while (v != r);
            }
            return;
        }
        for (auto u: neigh[v])
            if (dist[v] + 1 < dist[u]) {
                dist[u] = dist[v] + 1; par[u] = v;
                Q.push_back(u);
            }
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int a, b; scanf("%d %d", &a, &b);
        neigh[a].insert(b);
    }
    for (int i = 1; i <= n; i++)
        Solve(i);
    if (best >= Maxn) printf("-1\n");
    else {
        printf("%d\n", best);
        for (int i = 0; i < bi.size(); i++)
            printf("%d\n", bi[i]);
    }
    return 0;
}