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;
struct Edge {
  int nxt, to;
} e[1000010];
int vis[300010], root = 0;
int dfs(int n, int fa) {
  set<int> s;
  for (int i = e[n].nxt; i != -1; i = e[i].nxt) {
    if (e[i].to == fa) continue;
    int ans = dfs(e[i].to, n);
    if (ans == -1) {
      return -1;
    } else
      s.insert(ans + 1);
  }
  if (s.size() == 0) return 0;
  if (s.size() == 1) {
    return *s.begin();
  }
  if (s.size() == 2 && fa == 0) {
    return *s.begin() + *s.rbegin();
  }
  if (s.size() == 2 && fa != 0) {
    root = n;
    return -1;
  }
  if (s.size() >= 3) {
    return -1;
  }
}
int main() {
  int n;
  int a, b;
  scanf("%d", &n);
  int k = n + 1;
  memset(e, -1, sizeof(e));
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[k].nxt = e[u].nxt;
    e[u].nxt = k;
    e[k].to = v;
    k++;
    e[k].nxt = e[v].nxt;
    e[v].nxt = k;
    e[k].to = u;
    k++;
  }
  int ans = dfs(1, 0);
  if (root && ans == -1) {
    ans = dfs(root, 0);
  }
  if (ans == -1)
    printf("%d\n", ans);
  else {
    while (!(ans % 2)) {
      ans = ans / 2;
    }
    printf("%d\n", ans);
  }
}
