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;

const int MAXN = 1010;

struct Edge
{
	int to, nxt;
}e[MAXN << 1];
int tot, h[MAXN];

inline void add(int u, int v)
{
	e[++tot] = (Edge){v, h[u]};
	h[u] = tot;
}

int n, m, d[MAXN], f[MAXN];
int cnt, num, ans[MAXN], now[MAXN];

void solve(int s)
{
	memset(d, 0, sizeof(d));
	memset(f, 0, sizeof(f));
	queue<int> q; q.push(s);
	while(!q.empty())
	{
		int x = q.front(); q.pop();
		for(int i = h[x]; i; i = e[i].nxt)
		{
			int y = e[i].to;
			if(!d[y])
			{
				f[y] = x;
				d[y] = 1;
				q.push(y);
			}
		}
		if(d[s]) break;
	}
	if(!d[s]) {num = MAXN; return;}
	now[num = 1] = s;
	int p = f[s];
	while(p && p != s)
	{
		now[++num] = p;
		p = f[p];
	}
}

int main()
{
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; ++i)
	{
		int u, v;
		scanf("%d %d", &u, &v);
		add(u, v);
	}
	cnt = MAXN;
	for(int i = 1; i <= n; ++i)
	{
		solve(i);
		if(num < cnt)
		{
			cnt = num;
			for(int i = 1; i <= cnt; ++i) ans[i] = now[i];
		}
	}
	if(cnt == MAXN) puts("-1");
	else
	{
		printf("%d\n", cnt);
		for(int i = 1; i <= cnt; ++i) printf("%d\n", ans[i]);
	}
	return 0;
}