N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
　　If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
　　Note that there could be more than one channel between two planets.

 

 

　　The input contains multiple cases.
　　Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
　　(2<=N<=200000, 1<=M<=1000000)
　　Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
　　A line with two integers '0' terminates the input.

 

 

　　For each case, output the minimal number of bridges after building a new channel in a line.

 

 

 

 

 

#pragma comment(linker, "/STACK:102400000,102400000")

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<queue>

using namespace std;

const int N = 2e6+, M = 2e6+, inf = 2e9, mod = 1e9+;

int n,m,dfn[N],low[N],top,vis[N],cnt,from,head[N],t,xx[N],yy[N],q[N],inq[N],mx,belong[N],scc,all;

struct ss{int to,next,id;}G[M];

void add(int u,int v) {G[t].next=head[u];G[t].to=v;G[t].id = ;head[u]=t++;}

void init()

{

    memset(head,-,sizeof(head));

    scc=;cnt=;top=;

    memset(dfn,,sizeof(dfn));

    memset(vis,,sizeof(vis));

    t=;

}

void dfs(int x,int fa) {

    dfn[x] = low[x] = ++cnt;

    q[++top] = x;

    inq[x]=;

    for(int i=head[x];i!=-;i=G[i].next) {

        int to = G[i].to;

        if(G[i].id) continue;

        G[i].id = G[i^].id=;

        if(!dfn[to]) {

            dfs(to,x);

            low[x] = min(low[x],low[to]);

        }

        else if(inq[to])low[x] = min(low[x],dfn[to]);

    }

    if(low[x]==dfn[x]) {

        scc++;

        do {

            inq[q[top]]=;

            belong[q[top]]=scc;

        }while(x!=q[top--]);

    }

}

void Tarjan() {

    for(int i=;i<=n;i++)

        if(!dfn[i]) dfs(i,-);

}

void rebuild()

{

    t=;

    memset(head,-,sizeof(head));

    for(int i=;i<=m;i++)

    {

        int x = xx[i];

        int y = yy[i];

        if(belong[x]==belong[y]) continue;

        add(belong[x],belong[y]);

        add(belong[y],belong[x]);

       // cout<<"dsadas "<<belong[x]<<" "<<belong[y]<<endl;

    }

}

void bfs(int x)

{

    memset(vis,,sizeof(vis));

    queue<int >Q;

    Q.push(x);

    vis[x]=;

    mx = ;

    while(!Q.empty())

    {

        int k = Q.front();

        Q.pop();

        for(int i=head[k];i!=-;i=G[i].next)

        {

            int to = G[i].to;

            if(vis[to]) continue;

            Q.push(to);

            vis[to] = vis[k] + ;

            if(vis[to]>mx) {

                mx = vis[to];

                from = to;

            }

        }

    }

}

int main()

{

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        if(n==&&m==) break;

        init();

        for(int i=;i<=m;i++)

        {

            int a,b;

            scanf("%d%d",&a,&b);

            add(a,b);add(b,a);

            xx[i]=a;

            yy[i]=b;

        }

        Tarjan();

        rebuild();

        mx=;from = ;

        bfs();

        bfs(from);

        printf("%d\n",scc-mx);

    }

    return ;

}

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