Problem: You are given an undirected unweighted tree consisting of n vertices.

An undirected tree is a connected undirected graph with n - 1 edges.

Your task is to choose two pairs of vertices of this tree (all the chosen vertices should be distinct) (x_1, y_1) and (x_2, y_2) in such a way that neither x_1 nor y_1 belong to the simple path from x_2 to y_2 and vice versa (neither x_2 nor y_2 should not belong to the simple path from x_1 to y_1).

It is guaranteed that it is possible to choose such pairs for the given tree.

Among all possible ways to choose such pairs you have to choose one with the maximum number of common vertices between paths from x_1 to y_1 and from x_2 to y_2. And among all such pairs you have to choose one with the maximum total length of these two paths.

It is guaranteed that the answer with at least two common vertices exists for the given tree.

The length of the path is the number of edges in it.

The simple path is the path that visits each vertex at most once.

Input

The first line contains an integer n — the number of vertices in the tree (6 ≤ n ≤ 2 ⋅ 10^5).

Each of the next n - 1 lines describes the edges of the tree.

Edge i is denoted by two integers u_i and v_i, the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i).

It is guaranteed that the given edges form a tree.

It is guaranteed that the answer with at least two common vertices exists for the given tree.

Output

Print any two pairs of vertices satisfying the conditions described in the problem statement.

It is guaranteed that it is possible to choose such pairs for the given tree.

Examples

Input

7
1 4
1 5
1 6
2 3
2 4
4 7


Output

3 6
7 5


Input

9
9 3
3 5
1 2
4 3
4 7
1 7
4 6
3 8


Output

2 9
6 8


Input

10
6 8
10 3
3 7
5 8
1 7
7 2
2 9
2 8
1 4


Output

10 6
4 5


Input

11
1 2
2 3
3 4
1 5
1 6
6 7
5 8
5 9
4 10
4 11


Output

9 11
8 10

Note

The picture corresponding to the first example: <image>

The intersection of two paths is 2 (vertices 1 and 4) and the total length is 4 + 3 = 7.

The picture corresponding to the second example: <image>

The intersection of two paths is 2 (vertices 3 and 4) and the total length is 5 + 3 = 8.

The picture corresponding to the third example: <image>

The intersection of two paths is 3 (vertices 2, 7 and 8) and the total length is 5 + 5 = 10.

The picture corresponding to the fourth example: <image>

The intersection of two paths is 5 (vertices 1, 2, 3, 4 and 5) and the total length is 6 + 6 = 12.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, const U &b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, const U &b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T1, class T2>
inline void gn(T1 &x1, T2 &x2) {
  gn(x1), gn(x2);
}
template <class T1, class T2, class T3>
inline void gn(T1 &x1, T2 &x2, T3 &x3) {
  gn(x1, x2), gn(x3);
}
template <class T1, class T2, class T3, class T4>
inline void gn(T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  gn(x1, x2), gn(x3, x4);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = (long long)a * a % m)
    if (b & 1) ans = (long long)ans * a % m;
  return ans;
}
int vst[200200];
int one[200200], two[200200], one_id[200200], two_id[200200];
int da[200200], db[200200], dp[200200];
vector<int> adj[200200];
void dfs0(int u, int fa = 0) {
  for (int i = adj[u].size(); i--;) {
    int v = adj[u][i];
    if (v != fa) {
      dfs0(v, u);
      if (vst[v]) vst[u] = 1;
    }
  }
}
void pre_dfs(int u, int fa = 0) {
  one[u] = 0;
  one_id[u] = u;
  two[u] = -1;
  for (int i = adj[u].size(); i--;) {
    int v = adj[u][i];
    if (v != fa && !vst[v]) {
      pre_dfs(v, u);
      if (one[v] + 1 > two[u]) {
        two[u] = one[v] + 1, two_id[u] = one_id[v];
        if (two[u] >= one[u]) {
          swap(one[u], two[u]);
          swap(one_id[u], two_id[u]);
        }
      }
    }
  }
}
queue<int> q;
void dfs(int u) {
  vst[u] = 1;
  for (int i = adj[u].size(); i--;) {
    int v = adj[u][i];
    if (!vst[v]) {
      dp[v] = dp[u] + 1;
      dfs(v);
    }
  }
}
int main() {
  int n;
  gn(n);
  for (int i = 1, u, v; i < n; i++) {
    gn(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (adj[i].size() > 2) vst[i] = 1;
  for (int i = 1; i <= n; i++)
    if (vst[i]) {
      dfs0(i);
      break;
    }
  for (int i = 1; i <= n; i++)
    if (vst[i]) pre_dfs(i);
  int ed = 1, A, B;
  for (int i = 1; i <= n; i++)
    if (adj[i].size() > 2) ed = i;
  {
    q = queue<int>();
    q.push(ed);
    for (int i = 1; i <= n; i++) da[i] = 0x3f3f3f3f;
    da[ed] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = adj[u].size(); i--;) {
        int v = adj[u][i];
        if (da[u] + 1 < da[v]) {
          da[v] = da[u] + 1;
          q.push(v);
        }
      }
    }
    int mx = -1, kkk = -5;
    for (int i = 1; i <= n; i++)
      if (vst[i]) {
        if (da[i] > mx) {
          mx = da[i];
          kkk = one[i] + two[i];
          ed = i;
        } else if (da[i] == mx) {
          if (one[i] + two[i] > kkk) {
            kkk = one[i] + two[i];
            ed = i;
          }
        }
      }
    A = ed;
  }
  {
    q = queue<int>();
    q.push(A);
    for (int i = 1; i <= n; i++) da[i] = 0x3f3f3f3f;
    da[A] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = adj[u].size(); i--;) {
        int v = adj[u][i];
        if (da[u] + 1 < da[v]) {
          da[v] = da[u] + 1;
          q.push(v);
        }
      }
    }
    int mx = -1, kkk = -5;
    for (int i = 1; i <= n; i++)
      if (vst[i]) {
        if (da[i] > mx) {
          mx = da[i];
          kkk = one[i] + two[i];
          ed = i;
        } else if (da[i] == mx) {
          if (one[i] + two[i] > kkk) {
            kkk = one[i] + two[i];
            ed = i;
          }
        }
      }
    B = ed;
  }
  cout << one_id[A] << ' ' << one_id[B] << endl;
  cout << two_id[A] << ' ' << two_id[B] << endl;
}