//以一号节点为根节点，求出所有节点到根结点的距离，以及所有点的子节点的个数

 //然后计算根据已知信息计算所有节点到当前结点的距离

 //然后扫描n个点，O(n)求解

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 struct node {

     int y, net;

 }e[maxn << ];

 int f[maxn], h[maxn];//h表示以当前节点的根结点的子节点的个数，f表示所有子节点到当前节点的距离和

 int n;

 int lin[maxn], len = ;

 int id;

 inline int read() {

     int x = , y = ;

     char ch = getchar();

     while(!isdigit(ch)) {

         if(ch == '-') y = -;

         ch = getchar();

     }

     while(isdigit(ch)) {

         x = (x << ) + (x << ) + ch - '';

         ch = getchar();

     }

     return x * y;

 }

 inline void insert(int xx, int yy) {

     e[++len].y = yy;

     e[len].net = lin[xx];

     lin[xx] = len;

 }

 int son_num(int x, int fa) {

     for(int i = lin[x]; i; i = e[i].net) {

         int to = e[i].y;

         if(to != fa) h[x] += son_num(to, x) + ;

     }

     return h[x];

 }

 void everyson_to_one_dis(int x, int fa, int z) {

     f[] += z;

     for(int i = lin[x]; i; i = e[i].net) {

         int to = e[i].y;

         if(to != fa) everyson_to_one_dis(to, x, z + );

     }

 }

 void everyone_to_x_dis(int x, int fa) {

     f[x] = f[fa] - (h[x] + ) + (n - h[x] - );

     for(int i = lin[x]; i; i = e[i].net) {

         int to = e[i].y;

         if(to != fa) everyone_to_x_dis(to, x);

     }

 }

 int main() {

     memset(lin, , sizeof(lin));

     n = read();

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

         int x, y;

         x = read(), y = read();

         insert(x, y);

         insert(y, x);

     }

     son_num(, );

     for(int i = lin[]; i; i = e[i].net)

         everyson_to_one_dis(e[i].y, , );

     for(int i = lin[]; i; i = e[i].net)

         everyone_to_x_dis(e[i].y, );

     id = ;

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

         if(f[id] > f[i]) id = i;

     cout << id << ' ' << f[id] << '\n';

     return ;

 }