#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
const ll mod = 998244353ll;
int n, m;
vector<vi> g, gt;
vi p, h;
set<int> par;
void dfs(int v, ll maxh) {
	par.insert(v);
	int upd = 0;
	for(auto i : g[v]) {
		if(maxh < h[i]) upd = 1;
		if(h[i]) maxh = max(maxh, h[i]);
	}
	for(auto i : gt[v]) {
		if(h[i]&&h[i]+1!=h[v]) maxh = max(maxh, h[i]);
	}
	for(auto i : g[v]) {
		if(!h[i]) continue;
		if(upd&&h[i] == maxh&& par.count(i)) {
			vi ans;
			while(v != i)
				ans.pb(v), v = p[v];
			ans.pb(i);
			cout << ans.size() << "\n";
			for(auto i : ans) cout << i << "\n";
			exit(0);
		}
	}
	for(auto i : g[v]) {
		if(h[i]) continue;
		h[i] = h[v] + 1;
		p[i] = v;
		dfs(i, maxh);
	}
	par.erase(v);
}

int main() {
	ios::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);
	cin >> n >> m;
	g.resize(n+1);
	gt.resize(n+1);
	for(int f, t, i = 0; i < m; i++) {
		cin >> f >> t;
		g[f].pb(t);
		gt[t].pb(f);
	}
	for(int i = 1; i <= n; i++) {
		p.assign(n+1, 0);
		h.assign(n+1, 0);
		h[i] = 1;
		dfs(i, 0);
	}
	cout << -1;
}

### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
const ll mod = 998244353ll;
int n, m;
vector<vi> g, gt;
vi p, h;
set<int> par;
void dfs(int v, ll maxh) {
	par.insert(v);
	int upd = 0;
	for(auto i : g[v]) {
		if(maxh < h[i]) upd = 1;
		if(h[i]) maxh = max(maxh, h[i]);
	}
	for(auto i : gt[v]) {
		if(h[i]&&h[i]+1!=h[v]) maxh = max(maxh, h[i]);
	}
	for(auto i : g[v]) {
		if(!h[i]) continue;
		if(upd&&h[i] == maxh&& par.count(i)) {
			vi ans;
			while(v != i)
				ans.pb(v), v = p[v];
			ans.pb(i);
			cout << ans.size() << "\n";
			for(auto i : ans) cout << i << "\n";
			exit(0);
		}
	}
	for(auto i : g[v]) {
		if(h[i]) continue;
		h[i] = h[v] + 1;
		p[i] = v;
		dfs(i, maxh);
	}
	par.erase(v);
}

int main() {
	ios::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);
	cin >> n >> m;
	g.resize(n+1);
	gt.resize(n+1);
	for(int f, t, i = 0; i < m; i++) {
		cin >> f >> t;
		g[f].pb(t);
		gt[t].pb(f);
	}
	for(int i = 1; i <= n; i++) {
		p.assign(n+1, 0);
		h.assign(n+1, 0);
		h[i] = 1;
		dfs(i, 0);
	}
	cout << -1;
}

```