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 <iostream>
#include <fstream>
#include <vector>
#include <stack>
#include <queue>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
typedef pair<int,int> PP;

/*
    freopen("input","r",stdin);
    freopen("output","w",stdout);
*/

int N;

vector<set<int> > G;
vector<set<int> > T;

int main() {
    ios::sync_with_stdio(false);
    cin >> N;
    G.resize(N);
    T.resize(N);
    for (int i = 1;i < N;i++) {
    	int a, b;
    	cin >> a >> b;
    	a--;
    	b--;
    	G[a].insert(b);
    	G[b].insert(a);
    }
    for (int i = 1;i < N;i++) {
    	int a, b;
    	cin >> a >> b;
    	a--;
    	b--;
    	T[a].insert(b);
    	T[b].insert(a);
    }
    queue<int> D;
    for (int i = 0;i < N;i++) {
    	if (G[i].size() == 1 && T[i].size() == 1) {
    		D.push(i);
    	}
    }
    int x = 0;
    while (D.size() > 0) {
    	int m = D.front();
    	D.pop();
    	x++;
    	set<int> f;
    	for (auto it : G[m]) {
    		G[it].erase(m);
    		f.insert(it);
    	}
    	for (auto it : T[m]) {
    		T[it].erase(m);
    		f.insert(it);
    	}
    	for (auto it : f) {
    		if (G[it].size() == 1 && T[it].size() == 1) {
    			D.push(it);
    		}
    	}
    }
    if (x == N) cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}