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;

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

const int N = 1e3 + 5;
vector<int> G[N], ans, cur;
bool vis[N];

void dfs(int u)
{
	vis[u] = 1;
	cur.push_back(u);
	for (int v : G[u])
		if (cur[0] == v && (!ans.size() || ans.size() > cur.size())) ans = cur;
		else if (!vis[v]) dfs(v);
	
	cur.pop_back();
}

int main()
{
	int n = read();
	int m = read();
	for (int i = 1; i <= m; ++i)
	{
		int u = read(), v = read();
		G[u].push_back(v);
	}
	
	for (int i = 1; i <= n; ++i) 
		memset(vis, 0, sizeof(vis)), dfs(i);
	
	if (ans.size() == 0) return puts("-1"), 0;
	printf("%d\n", ans.size());
	for (int i : ans) printf("%d\n", i);
	
	return 0;
}