/*
  tarjan缩点后，形成一棵树，求出叶子节点数tot，答案是(tot+1)/2
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#define N 1010
using namespace std;
int num[N],low[N],instack[N],vis[N],s[N],in[N],belong[N],top,cnt,indexx;
int head[N],n,m;
struct node
{
    int v,pre;
};node e[N*2];
void add(int i,int x,int y)
{
    e[i].v=y;
    e[i].pre=head[x];
    head[x]=i;
}
void tarjan(int u,int fa)
{
    num[u]=low[u]=++indexx;
    vis[u]=instack[u]=1;
    s[++top]=u;
    for(int i=head[u];i!=-1;i=e[i].pre)
      if((fa^1)!=i)
      {
          int v=e[i].v;
          if(!vis[v])
          {
              tarjan(v,i);
              low[u]=min(low[v],num[u]);
          }
          else if(instack[v])
            low[u]=min(low[v],low[u]);
      }
    int x;
    if(num[u]==low[u])
    {
        ++cnt;
        do
        {
            x=s[top--];
            instack[x]=0;
            belong[x]=cnt;
        }while(u!=x);
    }
}
void work()
{
    for(int i=1;i<=m;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        add(i*2-2,x,y);add(i*2-1,y,x);
    }
    for(int i=1;i<=n;i++)
      if(!vis[i])tarjan(i,-1);
    for(int i=1;i<=n;i++)
      for(int j=head[i];j!=-1;j=e[j].pre)
        if(belong[i]!=belong[e[j].v])
          in[belong[e[j].v]]++;
    int tot=0;
    for(int i=1;i<=cnt;i++)
      if(in[i]==1)tot++;
    printf("%d\n",(tot+1)/2);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        top=cnt=indexx=0;
        memset(num,0,sizeof(num));
        memset(low,0,sizeof(low));
        memset(instack,0,sizeof(instack));
        memset(vis,0,sizeof(vis));
        memset(s,0,sizeof(s));
        memset(head,-1,sizeof(head));
        memset(e,0,sizeof(e));
        memset(in,0,sizeof(in));
        memset(belong,0,sizeof(belong));
        work();
    }
    return 0;
}