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;
using Int = long long;
int main() {
    int N, M; cin >> N >> M;
    vector<vector<int>> D(N, vector<int>(N, 1e9));
    for (int i = 0; i < M; i++) {
        int A, B; cin >> A >> B; A--, B--;
        D[A][B] = 1;
    }
    for (int k = 0; k < N; k++) {
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                D[i][j] = min(D[i][j], D[i][k] + D[k][j]);
            }
        }
    }
    int start = -1, len = 1e9;
    for (int i = 0; i < N; i++) {
        if (D[i][i] < len) {
            start = i, len = D[i][i];
        }
    }
    if (start == -1) {
        cout << -1 << '\n';
    } else {
        cout << len << '\n';
        int node = start;
        do {
            cout << node + 1 << '\n';
            for (int j = 0; j < N; j++) {
                if (D[node][j] == 1 && D[j][start] == len - 1) {
                    node = j;
                    break;
                }
            }
        } while (--len);
    }
    return 0;
}