Codeforces Round #397 E. Tree Folding(脑洞)

Vanya wants to minimize a tree. He can perform the following operation multiple times: choose a vertex v, and two disjoint (except for v) paths of equal length a0 = v, a1, ..., ak, and b0 = v, b1, ..., bk. Additionally, vertices a1, ..., ak, b1, ..., bk must not have any neighbours in the tree other than adjacent vertices of corresponding paths. After that, one of the paths may be merged into the other, that is, the vertices b1, ..., bk can be effectively erased:

Help Vanya determine if it possible to make the tree into a path via a sequence of described operations, and if the answer is positive, also determine the shortest length of such path.

Input

The first line of input contains the number of vertices n (2 ≤ n ≤ 2·105).

Next n - 1 lines describe edges of the tree. Each of these lines contains two space-separated integers u and v (1 ≤ u, v ≤ n, u ≠ v) — indices of endpoints of the corresponding edge. It is guaranteed that the given graph is a tree.

Output

If it is impossible to obtain a path, print -1. Otherwise, print the minimum number of edges in a possible path.

Examples

input

6
1 2
2 3
2 4
4 5
1 6

output

3

input

7
1 2
1 3
3 4
1 5
5 6
6 7

output

-1

Note

In the first sample case, a path of three edges is obtained after merging paths 2 - 1 - 6 and 2 - 4 - 5.

It is impossible to perform any operation in the second sample case. For example, it is impossible to merge paths 1 - 3 - 4 and 1 - 5 - 6, since vertex 6 additionally has a neighbour 7 that is not present in the corresponding path.

分析：首先，假如我们一开始选择的根节点不在直径上的话，无论我们怎么折，最长链的长度一定都不变，然后我们就试着把根设定在最长链的中点，

这样一路合并下来，最后判断一下根节点往下链的深度一共有几种，大于等于三种的都是不合法，否则就将根节点的所有链都合并。

#include <bits/stdc++.h>
#define MOD 10000007
#define N 200005
using namespace std;
int n,u,v,l,ans,root,Max,deep[N],s[N],a[N];
set<int> f;
vector<int> G[N];
int dfs(int u,int fa,int Deep)
{
    int temp = u;
    s[Deep] = u;
    for(int i = 0;i < G[u].size();i++)
    {
        int v = G[u][i];
        if(v != fa) dfs(v,u,Deep+1);
    }
    if(Deep > Max)
    {
        Max = Deep;
        l = u;
        root = s[(Deep+1)/2];
    }
}
bool dfs2(int u,int fa)
{
    deep[u] = 1;
    for(int i = 0;i < G[u].size();i++)
    {
        int v = G[u][i];
        if(v != fa)
        {
            if(!dfs2(v,u)) return false;
            if(deep[u] == 1) deep[u] = 1 + deep[v];
            else
             if(u != root && deep[v] + 1 != deep[u]) return false;
            if(u == root && deep[v] + 1 != deep[u]) f.insert(deep[v] + 1);
        }
    }
    if(u == root)
    {
        if(f.size() > 1) return false;
        if(f.size()) ans = deep[u] - 1 + *f.begin();
        else ans = deep[u];
        while(ans != 1 && ans & 1) ans -= (ans>>1);
    }
    return true;
}
int main()
{
    scanf("%d",&n);
    for(int i = 1;i < n;i++)
    {
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1,0,1);
    dfs(l,0,1);
    if(!dfs2(root,0)) cout<<-1<<endl;
    else cout<<ans-1<<endl;
}