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:
// F.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 500500;

int n, m, dfn[MAX_N], level[MAX_N], low[MAX_N];
vector<int> G[MAX_N];

void dfs(int v, int l = 0)
{
    dfn[v] = 1, level[v] = l;
    low[l] = v;
    int pt = -1;
    for (int i = 0, siz = G[v].size(); i < siz; i++)
    {
        int to = G[v][i];
        if (dfn[to] == 1 && (pt == -1 || level[pt] < level[to]))
            pt = to;
    }

    if (pt != -1)
    {
        set<int> ret;
        ret.insert(pt);
        for (int i = l; low[i] != pt; i--)
            ret.insert(low[i]);
        int cnt = 0;
        for (int u : ret)
            for (int to : G[u])
                if (ret.count(to))
                    cnt++;

        if (cnt == ret.size())
        {
            cout << cnt << endl;
            for (int u : ret)
                cout << u << endl;
            exit(0);
        }
    }

    for (int i = 0, siz = G[v].size(); i < siz; i++)
    {
        int to = G[v][i];
        if (dfn[to] == 0)
            dfs(to, l + 1);
    }
    dfn[v] = 0;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1, u, v; i <= m; i++)
        scanf("%d%d", &u, &v), G[u].push_back(v);
    for (int i = 1; i <= n; i++)
        dfs(i, 0);
    puts("-1");
    return 0;
}