10
6 8
3 10
1 9
6 1
8 2
5 7
3 6
10 4
3 5
 1
 5
1 2
2 3
3 4
4 5
 7
2 3
2 6
1 2
3 7
6 5
3 4
 10
8 4
8 10
8 1
8 2
5 9
5 3
5 7
8 5
5 6
 10
8 6
3 4
10 9
7 10
1 3
4 2
6 1
5 8
9 5
 10
1 10
6 9
5 4
1 3
1 2
3 8
6 7
8 6
1 5
 20
13 4
7 6
9 13
9 2
9 16
4 18
16 5
11 15
16 14
13 12
13 8
16 17
7 1
9 19
17 3
15 20
1 11
2 7
5 10
 3
1 2
1 3
 50
18 47
18 12
42 16
21 44
43 39
20 27
25 1
11 29
9 24
19 42
3 34
16 38
14 48
23 41
8 43
29 2
40 4
45 36
43 9
49 3
41 17
41 15
35 49
43 20
5 32
34 46
8 21
43 45
29 40
39 26
27 33
24 50
38 7
39 19
43 31
24 10
3 14
39 22
43 23
47 13
20 11
25 8
16 30
44 18
19 37
39 5
49 28
25 35
46 6
 15
11 2
5 13
9 4
5 7
11 5
13 14
2 10
3 11
7 1
7 6
10 15
9 12
2 9
10 8
 31
15 2
23 19
11 4
26 6
1 20
14 11
19 13
26 12
12 27
20 14
22 21
4 16
10 18
5 23
4 31
12 8
2 17
19 30
14 26
2 28
22 7
23 9
20 5
5 15
9 29
15 10
6 25
6 24
9 3
11 22
 63
34 40
43 62
33 43
49 6
17 47
26 13
35 15
40 16
17 44
37 34
18 55
57 10
16 14
39 56
36 5
4 33
4 59
54 24
37 45
21 31
57 2
19 8
43 41
49 48
15 19
35 4
15 63
30 50
11 30
59 18
20 46
18 1
33 17
31 22
40 54
63 49
29 25
46 9
16 3
42 23
59 7
54 53
42 38
8 60
8 58
7 51
36 28
63 36
21 57
25 37
45 20
25 35
19 26
20 42
30 32
11 39
7 12
31 52
45 21
26 27
34 11
46 61

9
 1
 1
 1
 1
 2
 10
 18
 2
 13
 1
 3
 1

#include <bits/stdc++.h>
using namespace std;
long long Pow(long long a, long long b, long long mod) {
  long long x = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) x = x * a % mod;
    a = a * a % mod;
  }
  return x;
}
namespace multihash {
const int D = 2, N = 1e5 + 5;
array<int, D> B[N], iB[N], M, Z;
array<int, D> operator+(array<int, D> lhs, const array<int, D> &rhs) {
  for (int i = 0; i < D; ++i) {
    lhs[i] += rhs[i];
    if (lhs[i] >= M[i]) lhs[i] -= M[i];
  }
  return lhs;
}
array<int, D> operator-(array<int, D> lhs, const array<int, D> &rhs) {
  for (int i = 0; i < D; ++i) {
    lhs[i] -= rhs[i];
    if (lhs[i] < 0) lhs[i] += M[i];
  }
  return lhs;
}
array<int, D> operator*(array<int, D> lhs, const array<int, D> &rhs) {
  for (int i = 0; i < D; ++i) lhs[i] = (long long)lhs[i] * rhs[i] % M[i];
  return lhs;
}
array<int, D> pow(array<int, D> lhs, long long b) {
  array<int, D> x;
  x.fill(1);
  for (; b > 0; b >>= 1) {
    if (b & 1) x = x * lhs;
    lhs = lhs * lhs;
  }
  return x;
}
array<int, D> to_mhash(int x) {
  array<int, D> m;
  m.fill(x);
  return m;
}
void init(const array<int, D> &b, const array<int, D> &m) {
  M = m, B[1] = b, Z.fill(0), B[0].fill(1), iB[0].fill(1);
  for (int i = 0; i < D; ++i) iB[1][i] = Pow(B[1][i], M[i] - 2, M[i]);
  for (int i = 2; i < N; ++i) {
    B[i] = B[i - 1] * B[1];
    iB[i] = iB[i - 1] * iB[1];
  }
}
}  // namespace multihash
namespace merkle_tree {
using namespace multihash;
void get(vector<array<int, D>> &v, bool sorted = false) {
  if (!sorted) sort(v.begin(), v.end() - 1);
  int n = v.size();
  for (int i = 1; i < n; ++i) v[i] = v[i - 1] + v[i] * B[i];
}
void rget(vector<array<int, D>> &v) {
  int n = v.size();
  for (int i = n - 1; i > 0; --i) v[i] = (v[i] - v[i - 1]) * iB[i];
}
int solve(const vector<vector<int>> &adj) {
  int n = adj.size(), pos;
  vector<int> sz(n, 1);
  vector<vector<array<int, D>>> h(n), ch(n), rh(n);
  function<void(int, int)> dfs1 = [&](int u, int p) {
    for (auto v : adj[u])
      if (v != p) {
        dfs1(v, u);
        sz[u] += sz[v];
        h[u].push_back(h[v].back());
      }
    h[u].push_back(to_mhash(sz[u]));
    get(h[u]);
  };
  dfs1(0, -1);
  auto f = [&](array<int, D> x) { return (long long)x[0] << 32 | x[1]; };
  map<long long, int> mp;
  for (int u = 0; u < n; ++u) ++mp[f(h[u].back())];
  int cnt = mp.size(), best = 0, node;
  auto update = [&](array<int, D> u, int c) {
    long long x = f(u);
    cnt -= mp[x] > 0;
    mp[x] += c;
    cnt += mp[x] > 0;
  };
  function<void(int, int)> dfs2 = [&](int u, int p) {
    if (cnt > best) best = cnt, node = u;
    rh[u] = h[u];
    rget(rh[u]);
    for (auto v : adj[u])
      if (v != p) {
        update(h[v].back(), -1);
        update(h[u].back(), -1);
        sz[u] -= sz[v];
        sz[v] = n;
        ch[v] = h[v];
        pos = lower_bound(rh[u].begin(), rh[u].end() - 1, h[v].back()) -
              rh[u].begin();
        h[u].back() = (pos ? h[u][pos - 1] : Z) +
                      (h[u].end()[-2] - h[u][pos]) * iB[1] +
                      to_mhash(sz[u]) * B[h[u].size() - 2];
        rget(h[v]);
        h[v].back().fill(sz[v]);
        h[v].insert(lower_bound(h[v].begin(), h[v].end() - 1, h[u].back()),
                    h[u].back());
        get(h[v], true);
        update(h[v].back(), +1);
        update(h[u].back(), +1);
        dfs2(v, u);
        update(h[v].back(), -1);
        update(h[u].back(), -1);
        h[u].back() = h[u].end()[-2] + rh[u].back() * B[h[u].size() - 1];
        h[v] = ch[v];
        sz[v] = n - sz[u];
        sz[u] = n;
        update(h[v].back(), +1);
        update(h[u].back(), +1);
      }
  };
  dfs2(0, -1);
  return node;
}
}  // namespace merkle_tree
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  multihash::init({13, 23}, {(int)1e9 + 7, (int)1e9 + 9});
  int n;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cout << merkle_tree::solve(adj) + 1 << "\n";
  return 0;
}
