題面：

B. Forming Teams time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

One day n students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.

We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student A is an archenemy to student B, then student B is an archenemy to student A.

The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.

Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.

Input

The first line contains two integers n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of students and the number of pairs of archenemies correspondingly.

Next m lines describe enmity between students. Each enmity is described as two numbers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.

You can consider the students indexed in some manner with distinct integers from 1 to n.

Output

Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.

Examples Input

5 4
1 2
2 4
5 3
1 4

Output

1

Input

6 2
1 4
3 4

Output

0

Input

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

Output

2

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題意：

    給定n個朋友，m個敵對關係。敵對關係是相互的，即a和b是敵對的，那麼b和a也是敵對的，具有敵對關係的兩個人不能在一個小組，一個人最多隻有兩個敵對物件。現在要求將所有人分成兩個小組，使得兩組人數相同，且內部不具備敵對關係，且做板凳（即未被選中的人的數量儘量少）。

解題：

    將題目給定的敵對關係構造成圖，可以發現，聯通塊要麼是環，要麼是鏈，要麼就是孤立點（因為一個點最多隻有2的度數）。分析可以比較容易得出，奇數長度的環，必須犧牲一個人做板凳，因為選出的len-1人分為兩個小組，必有一人與兩個小組的人都存在敵對關係，故而找奇環，發現則做板凳人數數量加一。如果是鏈，可以i,i+1分別分配給兩個小組，就不會出現敵對關係，但奇數鏈會損失一個，但不同鏈之間無敵對關係，因此可以連結所有鏈，最終鏈長度為奇數，則損失一個。

     注意：找奇環的過程，開始寫錯了，應該是計算一個聯通塊上的總度數，並記錄是否出現了度數為1或者為0的點，沒有則判定為環，奇環的總度數/2是奇數，可以通過這點判斷奇環。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
bool vis[105];
bool con[105][105];
int degree[105];
int minuss=0,n,m,sum;
bool flag;
void dfs(int x,int pre)
{
	vis[x]=1;
	if(degree[x]==1||degree[x]==0)
		flag=1;
	sum+=degree[x];
    for(int i=1;i<=n;i++)
	{
		if(!vis[i]&&con[x][i]&&i!=pre)
        {
			dfs(i,x);
		}
	}
}
int main()
{
	int a,b,ans;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&a,&b);
		degree[a]++;
		degree[b]++;
		con[a][b]=con[b][a]=1;
	}
	for(int i=1;i<=n;i++)
	{
		if(!vis[i])
        {
			sum=0;
			flag=0;
			dfs(i,-1);
			if(!flag)
		    {
				if(sum/2%2)
				minuss++;
			}
		}
	}
	ans=minuss+(n-minuss)%2;
	printf("%d\n",ans);
	return 0;
}