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:
#pragma GCC optimize("Ofast")
#include "bits/stdc++.h"

using namespace std;

typedef long int li;
typedef long double ld;
typedef long long ll;
typedef pair<li, li> pt;

#define fi first
#define se second
#define lson ((p<<1))
#define rson ((p<<1)+1)
#define mid ((L+R)>>1)
#define dump(x) cout << #x << " = " << x << endl
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

ll const INF = 1e12;

ll n, m, start;
bitset<1007> vis;
vector<li> g[1007], res, now;

void dfs(li v) {
    vis[v] = true;

    now.push_back(v);

    for(auto i : g[v]) {
        if(i == start && v != start && (res.size() == 0 || res.size() > now.size())) res = now;
        if(!vis[i]) dfs(i);
    }

    now.pop_back();
}

int main() {
    fastIO;
    cin >> n >> m;
    for(li i = 0; i < m; i ++) {
        li u, v;
        cin >> u >> v;
        g[--u].push_back(--v);
    }

    for(li i = 0; i < n; i ++) {
        start = i;
        dfs(i);
        vis.reset();
    }

    if(res.size() == 0) cout << -1;
    else {
        cout << res.size() << "\n";
        for(auto i : res) cout << i + 1 << "\n";
    }
}