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 N = 1003;

int n, m;
vector<int> a[N];

vector<int> ans;

bool c[N];
int p[N];
void bfs(int xx)
{
    memset(c, false, sizeof c);
    queue<int> q;
    q.push(xx);
    c[xx] = true;
    while (!q.empty())
    {
        int x = q.front();
        q.pop();
        for (int i = 0; i < a[x].size(); ++i)
        {
            int h = a[x][i];
            if (!c[h])
            {
                c[h] = true;
                p[h] = x;
                q.push(h);
            }
            if (h == xx)
            {
                vector<int> v;
                for (int y = x; y != h; y = p[y])
                    v.push_back(y);
                v.push_back(h);
                if (ans.empty() || v.size() < ans.size())
                    ans = v;
                return;
            }
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    while (m--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        a[x].push_back(y);
    }
    for (int i = 1; i <= n; ++i)
    {
        bfs(i);
    }
    if (ans.empty())
        cout << -1 << endl;
    else
    {
        cout << ans.size() << endl;
        for (int i = 0; i < ans.size(); ++i)
            cout << ans[i] << endl;
    }
    return 0;
}