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 FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

constexpr ll mod = 1e9 + 7;

template <typename T> inline bool chmax(T &a, const T b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> inline bool chmin(T &a, const T b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int n, m;
vector<int> e[1000];

bool used[1000];
vector<int> ans, v;
void dfs(int x) {
    v.push_back(x);
    used[x] = true;
    for(int y : e[x]) {
        if(v[0] == y)
            if(ans.empty() || ans.size() > v.size()) ans = v;
        if(!used[y]) dfs(y);
    }
    v.pop_back();
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> n >> m;
    rep(i, m) {
        int u, v;
        cin >> u >> v;
        e[u - 1].push_back(v - 1);
    }

    rep(i, n) {
        rep(i, n) used[i] = false;
        dfs(i);
    }

    if(ans.empty())
        cout << -1 << endl;
    else {
        cout << ans.size() << endl;
        for(int x : ans)
            cout << x + 1 << endl;
    }

    return 0;
}