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 <iostream>
#include <cstdio>
#include <vector>

using namespace std;

typedef long long ll;

const ll M = 4e18;

ll n, m, mn = M;
vector<ll> ve[1005];
vector<ll> ans;
vector<ll> now;
ll vv[1005];

void dfs(ll p)
{
    if(now.size() != 0 && p == now[0])
    {
        if(mn > now.size())
        {
            mn = now.size();
            ans = now;
        }
        return;
    }
    else if(vv[p]) return;
    if(now.size() > mn) return;
    vv[p] = true;
    now.push_back(p);
    for(ll i = 0; i < ve[p].size(); i++)
    {
        dfs(ve[p][i]);
    }
//    vv[p] = false;
    now.pop_back();
}

int main()
{
    scanf("%lld%lld", &n, &m);
    for(ll i = 0; i < m; i++)
    {
        ll u, v;
        scanf("%lld%lld", &u, &v);
        ve[u].push_back(v);
    }
    for(ll i = 1; i <= n; i++)
    {
        for(ll i = 1; i <= n; i++) vv[i] = false;
        dfs(i);
    }
    if(mn == M) printf("-1\n");
    else
    {
        printf("%lld\n", mn);
        for(ll i = 0; i < ans.size(); i++)
        {
            printf("%lld\n", ans[i]);
        }
    }
    return 0;
}