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;
#define all(v) ((v).begin()),((v).end())
#define ll long long
#define fastIO cout << fixed << setprecision(12), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
double const EPS = 1e-10, PI = acos(-1);
const int N = 3e5 + 9, M = 1e5 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const ll inf = 1e18;

vector<int> adj[N];
int vis[N];
int len = OO;
vector<int> vv, ans;

void cycleAns(int node, int root, bool awlMra) {
  if(!awlMra && root == node) {
    if((int)vv.size() < len) {
      len = vv.size();
      ans = vv;
    }
    return;
  }
  if(vis[node])
    return;
  vis[node] = 1;
  vv.push_back(node);
  for(auto it: adj[node])
    cycleAns(it, root, 0);
  vis[node] = 2;
  vv.pop_back();
}

int main() {
  fastIO;
#ifdef LOCAL
  freopen("input.in", "rt", stdin);
#endif
  int n, m, u, v;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
  }
  for (int i = 1; i <= n; ++i) {
    memset(vis, 0, sizeof vis);
    vv.clear();
    cycleAns(i, i, 1);
  }
  if(len > n) len = -1;
  cout << len;
  if(len != -1) {
    cout << '\n';
    for(int it: ans)
      cout << it << '\n';
  }
  return 0;
}