#include <bits/stdc++.h>

using namespace std;

// BFS-based 2-coloring of the graph
bool canBeBipartite(int start, const vector<vector<int>>& graph,
                    vector<int>& color)
{
	queue<int> q;
	q.push(start);
	color[start] = 1; // Assign color 1 (black) to the starting node

	while (!q.empty())
	{
		int node = q.front();
		q.pop();

		for (int neighbor : graph[node])
		{
			if (color[neighbor] == 0)
			{
				// If the neighbor has not been colored, color it with the opposite color
				color[neighbor] = -color[node];
				q.push(neighbor);
			}
			else if (color[neighbor] == color[node])
			{
				// If the neighbor has the same color as the current node, it's a conflict
				return false;
			}
		}
	}

	return true; // No conflicts found, the graph can be colored as a bipartite graph
}

int minControlsToDisconnect(int n, int m, const vector<pair<int, int>>& edges)
{
	vector<vector<int>> graph(n +
	                          1); // Use 1-based indexing for convenience with input
	for (const auto& edge : edges)
	{
		graph[edge.first].push_back(edge.second);
		graph[edge.second].push_back(edge.first);
	}

	vector<int> color(n + 1,
	                  0); // 0 means not colored, 1 means black, -1 means white

	for (int i = 1; i <= n; ++i)
	{
		if (color[i] == 0 && !canBeBipartite(i, graph, color))
		{
			// If we find a node that cannot be colored without conflicts,
			// it means the graph is not bipartite and we cannot disconnect all edges by controlling some nodes
			return -1;
		}
	}

	// Count the number of nodes colored with color 1 (black)
	int controls = count(color.begin() + 1, color.end(), 1);

	// Since the graph is bipartite, controlling all nodes of one color will disconnect all edges
	// (assuming we have a valid bipartite coloring, which we checked above)
	return min(controls, n -
	           controls); // Return the smaller set of nodes to control
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n, m;
	cin >> n >> m;

	vector<pair<int, int>> edges(m);
	for (int i = 0; i < m; ++i)
	{
		cin >> edges[i].first >> edges[i].second;
	}

	int result = minControlsToDisconnect(n, m, edges);
	if(result < 0)
	{
		cout << "Impossible";
	}
	else
	{
		cout << result;
	}
	return 0;
}

救命，测试点②③④⑤⑦全部答案错误了

稻花香里说丰年，听取 $\text{\color{red}{Wrong Answer}}$ 声一片（