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;
typedef long long ll;
const int N = 1e3+5;
int n, m;
vector<int> g[N];
int vis[N];
vector<int> tmp, ans;
void dfs(int u) {
    tmp.push_back(u);
    vis[u] = 1;
    for(int v : g[u]) {
        if(v == tmp[0]) {
            if(tmp.size() < ans.size() || ans.empty()) {
                ans = tmp;
            }
        } else if(!vis[v]) {
            dfs(v);
        }
    }
    tmp.pop_back();
}

int main() {
    cin >> n >> m;
    for(int i = 0, u, v; i < m; i++) {
        cin >> u >> v;
        g[u].push_back(v);
    }
    for(int i = 1; i <= n; i++) {
        memset(vis, 0, sizeof vis);
        tmp.clear();
        dfs(i);
    }
    if(ans.size() == 0) cout << -1 << endl;
    else {
        cout << ans.size() << '\n';
        for(int x : ans) {
            cout << x << '\n';
        }
    }
    return 0;
}