/*

     题意：求一个点为根节点，使得到其他所有点的距离最短，是有向边，反向的距离+1

     树形DP：首先假设1为根节点，自下而上计算dp[1](根节点到其他点的距离)，然后再从1开始，自上而下计算dp[v],

             此时可以从上个节点的信息递推出来

 */

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <vector>

 using namespace std;

 const int MAXN = 2e5 + ;

 const int INF = 0x3f3f3f3f;

 struct Edge    {

     int v, w;

 };

 vector<Edge> G[MAXN];

 int dp[MAXN];

 bool vis[MAXN];

 int n, res;

 void DFS(int u)    {

     vis[u] = true;

     for (int i=; i<G[u].size (); ++i)    {

         int v = G[u][i].v, w = G[u][i].w;

         if (vis[v])    continue;

         DFS (v);

         res += w;

     }

 }

 void DFS2(int u)    {

     vis[u] = true;

     for (int i=; i<G[u].size (); ++i)    {

         int v = G[u][i].v, w = G[u][i].w;

         if (vis[v])    continue;

         if (w == )    dp[v] = dp[u] + ;

         else    dp[v] = dp[u] - ;

         DFS2 (v);

     }

 }

 void work(void)    {

     memset (vis, false, sizeof (vis));

     memset (dp, , sizeof (dp));

     res = ;    DFS ();    dp[] = res;

     memset (vis, false, sizeof (vis));

     DFS2 ();

     int mn = INF, p = ;

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

         if (mn >= dp[i])    {

             mn = dp[i];    p = i;

         }

     }

     printf ("%d\n", mn);

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

         if (i == p)    {

             printf ("%d\n", i);    break;

         }

         if (dp[i] == mn)    printf ("%d ", i);

     }

 }

 int main(void)    {        //Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland

 //    freopen ("A.in", "r", stdin);

     while (scanf ("%d", &n) == )    {

         for (int i=; i<=n; ++i)    G[i].clear ();

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

             int u, v;    scanf ("%d%d", &u, &v);

             G[u].push_back ((Edge) {v, });

             G[v].push_back ((Edge) {u, });

         }

         work ();

     }

     return ;

 }