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>
#define rep(i, n) for (int i = 0; i < n; ++i)
using namespace std;

const int inf = 1e9;
int G[1000][1000];
int D[1000][1000];
int main() {
  int n, m;
  cin >> n >> m;
  rep(i, n) {
    rep(j, n) {
      G[i][j] = D[i][j] = inf;
    }
  }
  
  rep(z, m) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    G[a][b] = D[a][b] = 1;
  }
  
  rep(k, n) rep(i, n) rep(j, n)
    D[i][j] = min(D[i][j], D[i][k] + D[k][j]);
  
  int l = inf;
  int s = 0;
  rep(i, n) {
    if (D[i][i] < l) {
      l = D[i][i];
      s = i;
    }
  }
  
  if (l == inf) {
    cout << -1 << endl; 
  } else {
    cout << l << endl;
    int c = s;
    int k = l;
    while (true) {
      --k;
      cout << c + 1 << endl;
      int nxt = -1;
      rep(i, n) {
        if (c == i) continue;
        if (G[c][i] == inf) continue;
        if (D[i][s] == k) {
          nxt = i;
          break;
        }
      }
      if (nxt == -1) break;
      c = nxt;
    }
  }
}