E. Bear and Drawing

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     E. Bear and Drawingtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Limak is a little bear who learns to draw. People usually start with houses, fences and flowers but why would bears do it? Limak lives in the forest and he decides to draw a tree.

Recall that tree is a connected graph consisting of n vertices and n - 1 edges.

Limak chose a tree with n vertices. He has infinite strip of paper with two parallel rows of dots. Little bear wants to assign vertices of a tree to some n distinct dots on a paper so that edges would intersect only at their endpoints — drawn tree must be planar. Below you can see one of correct drawings for the first sample test.

Is it possible for Limak to draw chosen tree?

Input

The first line contains single integer n (1 ≤ n ≤ 105).

Next n - 1 lines contain description of a tree. i-th of them contains two space-separated integers ai and bi(1 ≤ ai, bi ≤ n, ai ≠ bi) denoting an edge between vertices ai and bi. It‘s guaranteed that given description forms a tree.

Output

Print "Yes" (without the quotes) if Limak can draw chosen tree. Otherwise, print "No" (without the quotes).

Sample test(s)input

8
1 2
1 3
1 6
6 4
6 7
6 5
7 8

output

Yes

input

13
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13

output

No

我們想如果一個點 串連著 大於等於兩顆樹是多叉樹的話，那麼我們就可以認為這個是無解的（除非其中一顆樹是直接連接在該結點上的）。

#include<cstdio>#include <iostream>#include <string.h>#include <vector>using namespace std;const int maxn = 100000 + 5;vector<int> G[maxn];int leg[maxn];bool del[maxn];void dfs(int cur, int per=-1){      if(G[cur].size()<=2)        {             del[cur]=true;             int siz=G[cur].size();             for(int i=0; i<siz; i++)                {                     int b=G[cur][i];                     if(b == per) continue;                     dfs(b,cur);                }        }}int main() {    int n;    while(scanf("%d",&n)==1)    {        memset(del,false,sizeof(del));        memset(leg,0,sizeof(leg));         for(int i=1; i<=n; i++)G[i].clear();        for(int i=1; i<n; i++)            {                int a,b;                 scanf("%d%d",&a,&b);                 G[a].push_back(b);                 G[b].push_back(a);            }            for(int i=1; i<=n; i++)            if(G[i].size() ==1 ){                dfs(i);            }            for(int a=1; a<=n; a++)            {                int siz=G[a].size();                for(int j = 0; j<siz; j++)                    {                         int b=G[a][j];                         if(del[b])                            {                                leg[a]=min(leg[a]+1,2);                            }                    }            }            bool falg=true;            for(int a=1; a<=n; a++)                if(del[a]==false){                     int cnt=0;                     for(int j=0; j<G[a].size(); j++)                        {                            int b=G[a][j];                           if(del[b]==false&&G[b].size()-leg[b]>1) cnt++;                        }                      if(cnt>2){                        falg=false; break;                      }                }                if(falg)puts("YES");                else puts("NO");    }    return 0;}

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E. Bear and Drawing