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 MAXN = 200005;
vector<int> gra[MAXN], da[MAXN];
int d[MAXN];
queue<int> q;
int main() {
  int n, u, v, temd, ans, restn;
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; ++i) {
    scanf("%d%d", &a, &b);
    gra[a].push_back(b);
    gra[b].push_back(a);
    d[a]++;
    d[b]++;
  }
  for (int i = 1; i <= n; ++i)
    if (d[i] == 1) q.push(i), da[i].push_back(0);
  restn = n;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    restn--;
    sort(da[u].begin(), da[u].end());
    da[u].erase(unique(da[u].begin(), da[u].end()), da[u].end());
    if (da[u].size() > 2 || da[u].size() == 2 && restn > 0) {
      u = -1;
      break;
    }
    temd = da[u].front() + 1;
    for (vector<int>::iterator iter = gra[u].begin(); iter != gra[u].end();
         ++iter)
      if (v = *iter, d[v] > 0) {
        da[v].push_back(temd);
        d[v]--;
        d[u]--;
        if (d[v] == 1) q.push(v);
      }
  }
  if (u < 0)
    printf("-1\n");
  else {
    ans = da[u][0] + da[u][1];
    while (ans % 2 == 0) ans /= 2;
    printf("%d\n", ans);
  }
  return 0;
}
