Description

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 isnot 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 consideredseparate, 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: thei-th road is determined by two distinct integersx_i,?y_i (1?≤?x_i,?y_i

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.

Sample Input

4 3
2 1
1 3
4 3

1

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

0

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

1

Hint

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

The second sample: ,,,,.

The third sample: ,,,,.

題意：給出一些點和邊。要求你把邊變成有向邊使得入度為0的點最少

分析：對於一些相連的點來說，假設形成的圖中有環。那麽我們一定可以從環上的某點出發使入度為0的點沒有，

假設無環，那麽僅僅須要一個入度為0的點就能使其它點入度不為0，那麽問題就轉換為推斷圖中是否有環了，

我們從任一個點遍歷整個圖，然後判環調整答案就可以

#include<cstring>
#include<string>
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>
//#pragma comment(linker, "/STACK:1024000000,1024000000");

using namespace std;

#define INF 0x3f3f3f3f

vector<int>v[100006];
int vis[100006];
int ans;
int flag;
void dfs(int s,int f,int pre)
{
    vis[s]=1;
    for(int i=0; i<v[s].size(); i++)
    {
        if(vis[v[s][i]]&&pre!=v[s][i]) flag=-1;
        else if(!vis[v[s][i]])
        {
            dfs(v[s][i],f,s);
        }
    }
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(vis,0,sizeof vis);
        for(int i=1; i<=n; i++) v[i].clear();
        for(int i=0; i<m; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            v[a].push_back(b);
            v[b].push_back(a);
        }
        ans=0;
        for(int i=1; i<=n; i++)
        {
            if(!vis[i])
            {
                flag=0;
                ans++;
                dfs(i,i,i);
                ans+=flag;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

CodeForces 659E New Reform （圖的遍歷判環）