10
4 2
7 4
2 6
2 5
4 8
10 3
2 9
9 1
5 10
 10
5 10
7 8
8 3
2 6
3 2
9 7
4 5
10 1
6 4
 4
4 1
4 3
4 2
 3
3 1
1 2
 9
1 2
1 3
1 4
1 5
1 6
6 7
6 8
8 9
 5
1 5
2 3
2 4
1 2
 8
6 1
4 7
4 8
8 5
7 6
4 3
4 2
 7
5 6
5 7
5 1
7 4
6 3
3 2
 3
1 3
3 2
 2
1 2
 5
5 4
4 3
3 1
5 2
 6
1 6
3 1
6 4
5 3
2 5
 11
8 9
2 7
1 11
3 2
9 1
8 5
8 6
5 4
4 10
8 3
 11
11 9
6 7
7 1
8 11
5 6
3 5
9 3
10 8
2 4
4 10
 12
12 6
6 7
8 11
4 8
10 4
12 3
2 10
6 2
12 9
4 1
9 5

-1
 9
 1
 1
 3
 3
 -1
 -1
 1
 1
 1
 5
 1
 5
 -1

#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
const double pi = acos(-1.0);
const int INF = INT_MAX;
const long long LLINF = LLONG_MAX;
const int MAX_N = 2e5 + 10;
int N, u, v, ans;
vector<int> G[MAX_N];
int dfs(int v, int p) {
  set<int> h;
  for (auto u : G[v])
    if (u != p) {
      int res = dfs(u, v);
      if (res == -1)
        return -1;
      else
        h.insert(res + 1);
    }
  if (!h.size()) return 0;
  if (h.size() == 1) return *h.begin();
  if (!p && h.size() == 2) return *h.begin() + *h.rbegin();
  u = v;
  return -1;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &N);
  for (int i = 1; i < N; ++i)
    scanf("%d%d", &u, &v), G[v].push_back(u), G[u].push_back(v);
  u = 0;
  ans = dfs(1, 0);
  if (u) ans = dfs(u, 0);
  while (~ans && !(ans & 1)) ans /= 2;
  printf("%d\n", ans);
  return 0;
}
