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, s, n)    for (int i = (s); i < (n); ++i)
using namespace std;

int V, E;
vector<int> G[1005];
enum {
    NEW, ACTIVE, FINISHED
} stat[1005];

bool dfs(int u) {
    stat[u] = ACTIVE;
    for (const int v : G[u]) {
        if (stat[v] == ACTIVE || dfs(v)) return true;
    }
    stat[u] = FINISHED;
    return false;
}

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);

    cin >> V >> E;
    rep(i, 0, E) {
        int a, b;
        cin >> a >> b;
        --a, --b;
        G[a].push_back(b);
    }

    rep(u, 0, V) {
        if (stat[u] == NEW && dfs(u)) {
            cout << 0 << endl;
            return 0;
        }
    }

    cout << -1 << endl;

    return 0;
}