/*

 tarjan+缩点+树的直径

 题意：给出n个点和m条边的图，存在重边，问加一条边以后，剩下的桥的数量最少为多少。

 先tarjan缩点，再在这棵树上求直径。加的边即是连接这条直径的两端。

 */

 #pragma comment(linker, "/STACK:1024000000,1024000000")

 #include<stdio.h>

 #include<string.h>

 #include<stdlib.h>

 #include<stack>

 #include<algorithm>

 #include<iostream>

 #include<queue>

 #include<map>

 #include<math.h>

 using namespace std;

 typedef long long ll;

 //typedef __int64 int64;

 const int maxn = ;

 const int maxm = ;

 const int inf = 0x7fffffff;

 const double pi=acos(-1.0);

 const double eps = 1e-;

 struct Edge{

     int v,next;

 }edge[ maxm<< ],edgeTree[ maxm<< ];

 int cnt,cnt2,head[ maxn ],head2[ maxn ];

 int vis[ maxn ];

 int dfn[ maxn ];

 int low[ maxn ];

 int be[ maxn ];//缩点

 int sum;//the num of "缩点"

 int id;

 stack<int>sta;

 queue<int>q;

 int maxNode,maxs;//这棵树的直径maxs

 int dis[ maxn ];

 map<int,int>mp;

 void init(){

     cnt = cnt2 = ;

     id = ;

     sum = ;

     //mp.clear();

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

     memset( head2,-,sizeof( head2 ) );

     memset( dfn,-,sizeof( dfn ) );

     memset( vis,-,sizeof( vis ) );

     memset( low,-,sizeof( low ) );

     while( !sta.empty() )

         sta.pop();

 }

 void addedge( int a,int b ){

     edge[ cnt ].v = b;

     edge[ cnt ].next = head[ a ];

     head[ a ] = cnt++;

 }

 void addedge2( int a,int b ){

     edgeTree[ cnt2 ].v = b;

     edgeTree[ cnt2 ].next = head2[ a ];

     head2[ a ] = cnt2++;

 }

 void tarjan( int u,int Link ){

     dfn[ u ] = low[ u ] = id++;

     vis[ u ] = ;

     sta.push( u );

     int cc = ;

     for( int i=head[u];i!=-;i=edge[i].next ){

         int v = edge[ i ].v;

         if( dfn[ v ]==- ){

             tarjan( v,u );

             low[ u ] = min( low[ u ],low[ v ] );

         }

         else if( Link==v ){

             if( cc ) low[ u ] = min( low[ u ],dfn[ v ] );

             cc++;

         }

         else if( vis[ v ]== ){

             low[ u ] = min( low[ u ],dfn[ v ] );

         }

     }

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

         sum++;

         int tt;

         while(  ){

             tt = sta.top();

             sta.pop();

             vis[ tt ] = ;

             be[ tt ] = sum;

             if( tt==u ) break;

         }

     }

 }

 void buildTree( int n ){

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

         for( int j=head[ i ];j!=-;j=edge[ j ].next ){

             if( be[i]!=be[edge[j].v] ){

                 addedge2( be[i],be[edge[j].v] );

             }

         }

     }

 }

 void bfs( int s,int n ){

     memset( vis,,sizeof( vis ) );

     vis[ s ] = ;

     while( !q.empty() )

         q.pop();

     q.push( s );

     dis[ s ] = ;

     maxs = ;

     while( !q.empty() ){

         int cur = q.front();

         q.pop();

         if( dis[ cur ]>=maxs ){

             maxs = dis[ cur ];

             maxNode = cur;

         }

         for( int i=head2[ cur ];i!=-;i=edgeTree[ i ].next ){

             int v = edgeTree[ i ].v;

             if( vis[ v ]== ) continue;

             vis[ v ] = ;

             dis[ v ] = dis[ cur ]+;

             q.push( v );

         }

     }

     return ;

 }

 int dfs(int u,int p)

 {

     int max1=,max2=;

     for (int i=head2[u];i!=-;i=edgeTree[i].next)

     {

         int v=edgeTree[i].v;

         if (v==p) continue;

         int tmp=dfs(v,u)+;

         if (max1<tmp) max2=max1,max1=tmp;

         else if (max2<tmp) max2=tmp;

     }

     maxs=max(maxs,max1+max2);

     return max1;

 } //dfs求树的直径 ok

 int main(){

     int n,m;

     while( scanf("%d%d",&n,&m)&&(n+m) ){

         int a,b;

         init();

         while( m-- ){

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

             if( a==b ) continue;

             //if( mp[min(a,b)] == max(a,b) ) continue;

             //mp[min(a,b)] = max(a,b);

             //不能这样去重边？？？

             addedge( a,b );

             addedge( b,a );

         }

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

             if( dfn[i]==- ){

                 tarjan( i,- );

             }

         }

         buildTree( n );

         bfs( ,sum );

         bfs( maxNode,sum );

         //printf("sum=%d, maxs=%d\n",sum,maxs);

         //maxs = 0;

         //dfs( 1,-1 );

         printf("%d\n",sum--maxs);

     }

     return ;

 }