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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define P pair<int, int>
#define mk_p make_pair
const int N = 1e5 + 11;
set<int> s[N];
map<P, int> mp;
int n, fa[N];
P dl[N];
int read(){
    int x = 0, fl = 1;
    char ch = getchar();
    while(ch > '9' || ch < '0')fl = (ch == '-' ? -1 : fl), ch = getchar();
    while(ch >= '0' && ch <= '9')x = 10 * x + ch - 48, ch = getchar();
    return x * fl;
}
int find(int x){
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}
int main(){
    freopen("agc014e.in", "r", stdin);
    freopen("agc014e.out", "w", stdout);
    n = read();
    int u, v;
    int hd = 0, tl = 0;
    for(int i = 1;i <= 2 * n - 2; i++){
        u = read(); v = read();
        s[u].insert(v); s[v].insert(u);
        mp[mk_p(u, v)]++; mp[mk_p(v, u)]++;
        if(mp[mk_p(u, v)] == 2)dl[++tl] = mk_p(u, v);
    }
    for(int i = 1;i <= n; i++)fa[i] = i;
    while(hd < tl){
        P now = dl[++hd];
        int u = now.first, v = now.second;
        u = find(u); v = find(v);
        if(u == v)continue;
        if(s[u].size() < s[v].size())swap(u, v);
        fa[v] = u;
        set<int>::iterator it;
        s[u].erase(v); s[v].erase(u);
        for(it = s[v].begin();it != s[v].end(); it++){
            int to = *it;
            mp[mk_p(v, to)] = mp[mk_p(to, v)] = 0;
            s[to].erase(v); s[to].insert(u); s[u].insert(to);
            mp[mk_p(to, u)]++; mp[mk_p(u, to)]++;
            if(mp[mk_p(to, u)] == 2)dl[++tl] = mk_p(to, u);
        }
    }
    int fl = 0;
    for(int i = 1;i <= n; i++)fl += fa[i] == i;
    puts(fl == 1 ? "YES" : "NO");
    return 0;
}