Problem:
Score : 1400 points

Problem Statement

There is a tree with N vertices numbered 1 through N.
The i-th of the N-1 edges connects vertices a_i and b_i.

Initially, each edge is painted blue.
Takahashi will convert this blue tree into a red tree, by performing the following operation N-1 times:

Select a simple path that consists of only blue edges, and remove one of those edges.

Then, span a new red edge between the two endpoints of the selected path.

His objective is to obtain a tree that has a red edge connecting vertices c_i and d_i, for each i.

Determine whether this is achievable.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i,b_i,c_i,d_i ≤ N

a_i ≠ b_i

c_i ≠ d_i

Both input graphs are trees.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_{N-1} d_{N-1}

Output

Print YES if the objective is achievable; print NO otherwise.

Sample Input 1

3
1 2
2 3
1 3
3 2

Sample Output 1

YES

The objective is achievable, as below:

First, select the path connecting vertices 1 and 3, and remove a blue edge 1-2. Then, span a new red edge 1-3.

Next, select the path connecting vertices 2 and 3, and remove a blue edge 2-3. Then, span a new red edge 2-3.

Sample Input 2

5
1 2
2 3
3 4
4 5
3 4
2 4
1 4
1 5

Sample Output 2

YES

Sample Input 3

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

Sample Output 3

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int n,t,w;
int d[200010];
pair<int,int> q[200010];
map<int,int> h[200010];

int main()
{int i,j,x,y,k;
 map<int,int>::iterator p,p1;
 
 scanf("%d",&n);
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]=1;
 	h[y][x]=1;
 	d[x]++;
 	d[y]++;
 }
 t=1;
 w=0;
 for(i=1;i<n;i++)
 {	scanf("%d%d",&x,&y);
 	h[x][y]++;
 	h[y][x]++;
 	d[x]++;
 	d[y]++;
 	if(h[x][y]==2) q[++w]=make_pair(x,y);
 }
 while(t<=w)
 {	x=q[t].first;
 	y=q[t].second;
 	if(d[x]<d[y]) swap(x,y);
 	h[x][y]-=2;
 	h[y][x]-=2;
 	for(p=h[y].begin();p!=h[y].end();p++)
 	{	k=(h[x][(*p).first]+=(*p).second);
 		h[(*p).first][x]+=(*p).second;
 		h[(*p).first][y]-=(*p).second;
 		if(k==2)
 			q[++w]=make_pair(x,(*p).first);
 	}
	d[x]+=d[y]-2;
 	t++;
 }
 printf("%s\n",t==n?"YES":"NO");
 return 0;
}