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;
const int N = 2e5 + 9;
int n, rt;
set<int> g[N];
set<int> gg[N];
int du[N], ans;
bool vis[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1, u, v; i < n; i++)
    cin >> u >> v, g[u].insert(v), g[v].insert(u), du[u]++, du[v]++;
  if (n == 2) {
    cout << 1 << endl;
    return 0;
  }
  ans = n - 1;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) q.push(i), gg[i].insert(0);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    rt = t;
    if (gg[t].size() == 1) {
      for (auto v : g[t]) {
        g[v].erase(t);
        if (gg[v].count((*gg[t].begin()) + 1))
          ans = ans - (*gg[t].begin()) - 1;
        else
          gg[v].insert((*gg[t].begin()) + 1);
        if (g[v].size() == 1) q.push(v);
      }
    }
  }
  for (int i = 1; i <= n; i++)
    if (gg[i].size() == 0 || gg[i].size() > 2) {
      cout << -1 << endl;
      return 0;
    }
  while (ans % 2 == 0) ans /= 2;
  cout << ans << endl;
  return 0;
}
