codeforces 659E New Reform

E. New Reform

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guaranteed that you can get from any city to any other one, using only the existing roads.

The President of Berland decided to make changes to the road system and instructed the Ministry of Transport to make this reform. Now, each road should be unidirectional (only lead from one city to another).

In order not to cause great resentment among residents, the reform needs to be conducted so that there can be as few separate cities as possible. A city is considered separate, if no road leads into it, while it is allowed to have roads leading from this city.

Help the Ministry of Transport to find the minimum possible number of separate cities after the reform.

Input

The first line of the input contains two positive integers, n and m — the number of the cities and the number of roads in Berland (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000).

Next m lines contain the descriptions of the roads: the i-th road is determined by two distinct integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi), where xi and yi are the numbers of the cities connected by the i-th road.

It is guaranteed that there is no more than one road between each pair of cities, but it is not guaranteed that from any city you can get to any other one, using only roads.

Output

Print a single integer — the minimum number of separated cities after the reform.

Examples

input

4 3
2 1
1 3
4 3

output

1

input

5 5
2 1
1 3
2 3
2 5
4 3

output

0

input

6 5
1 2
2 3
4 5
4 6
5 6

output

1

Note

In the first sample the following road orientation is allowed: , , .

The second sample: , , , , .

The third sample: , , , , .

只要有一个环的连通分支，那么肯定可以做到环内的点都有入度，对于环外的点可以看成DFS遍历的生成树的节点，肯定可以有一条有向的通路，

通路上的点都有入度，这个连通分支可以做到最少为0个没有入度的点

如果没有环的连通分支，那么只需要找一个点作为生成树的根节点，那么就可以做到最少只有1个点没有入度

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>

using namespace std;

const int maxn = 100000 + 10;
vector<int> G[maxn];
bool vis[maxn], flag;
int n, m, ans;

void dfs(int s, int pre) {
	vis[s] = true;
	for (int i = 0; i < G[s].size(); i++) {
		if (vis[G[s][i]] && G[s][i] != pre) {
			flag = true;
		}
		else if (!vis[G[s][i]]) {
			dfs(G[s][i], s);
		}
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	memset(vis, false, sizeof(vis));
	ans = 0;
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {
			flag = false;
			dfs(i, i);
			if (!flag) ans++;
		}
	}
	printf("%d\n", ans);
	return 0;
}