Codeforces Round #397 E. Tree Fold(bfs,想法题,好题)

题目链接

E. Tree Folding

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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

61 22 32 44 51 6

output

3

input

71 21 33 41 55 66 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.

题意：

给一棵树，如果两条链，像上面一样，a[i]和b[i]都没有邻居，就可以合并它们，问最后能不能合并成一条链，且要求链长度最小

题解：

从叶子节点bfs,对于每一个节点用set保存从他下方的节点到这个节点的链的长度，如果有相同长度的就会自动“合并”（set的功能）。边bfs边删边，不断取叶子结点进行向上缩。

详见代码

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>#include<set>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=2e5+100;set<int> g[maxn];set<int> d[maxn];queue<int> q;int vis[maxn];int re(int x){    while(x%2==0)        x/=2;    return x;}int solve(){    while(!q.empty())    {        int u=q.front();        q.pop();        if(vis[u]) continue;        if(g[u].size()==1)        {            if(d[u].size()==1)            {                int v=*g[u].begin();                if(vis[v]) continue;                g[u].erase(v),g[v].erase(u);                d[v].insert(*d[u].begin()+1);                if(g[v].size()<=1)                q.push(v);                vis[u]=1;            }        }        else if(g[u].size()==0)        {            if(d[u].size()>=3) return -1;            else if(d[u].size()==2)                return re(*d[u].begin()+*d[u].rbegin());            else return re(*d[u].begin());        }    }    return -1;}int main(){    int n;    scanf("%d",&n);    rep(i,1,n)    {        int u,v;        scanf("%d%d",&u,&v);        g[u].insert(v),g[v].insert(u);    }    rep(i,1,n+1) if(g[i].size()==1) q.push(i),d[i].insert(0);    memset(vis,0,sizeof(vis));    printf("%d\n",solve());    return 0;}