#include <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int , int >
#define F first
#define S second
#define endl '\n'
#define int long long
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#define kill(x) return cout<<x<<'\n', 0;
using namespace std;
const int N=2e5+100;
vector <int> g[N],mol[N];
map <pii,int> yal;
queue <pii> q;
ll par[N];
void add_edge(ll u,ll v){
    yal[{u,v}]++;
    yal[{v,u}]++;
    if (yal[{u,v}]>1 && u!=v){
        q.push({u,v});
    }
    return ;
}
void mrg(ll u,ll v){
    if (mol[u].size()>mol[v].size()) swap(u,v);
    for (auto t : g[u]){
        add_edge(v,par[t]);
        g[v].pb(par[t]);
    }
    g[u].clear();
    for (auto t : mol[u]){
        par[t]=v;
        mol[v].pb(t);
    }
    mol[u].clear();
    return ;
}
int32_t main(){
    sync;
    ll n;
    cin >> n;
    for (int i=1;i<N;i++) par[i]=i,mol[i].pb(i);
    for (int i=1;i<2*n-1;i++){
        ll u,v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
        add_edge(u,v);
    }
    while(q.size()){
        pii a=q.front();
        q.pop();
       // cout << a.F << " " << a.S << endl;
        mrg(a.F,a.S);
    }
    ll y=par[1];
    for (int i=2;i<=n;i++){
        if (par[i]!=y) kill("NO");
    }
    kill("YES");

}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

Examples

Input

3
1 2
2 3
1 3
3 2


Output

YES


Input

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


Output

YES


Input

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


Output

NO

### Response

```cpp
#include <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int , int >
#define F first
#define S second
#define endl '\n'
#define int long long
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#define kill(x) return cout<<x<<'\n', 0;
using namespace std;
const int N=2e5+100;
vector <int> g[N],mol[N];
map <pii,int> yal;
queue <pii> q;
ll par[N];
void add_edge(ll u,ll v){
    yal[{u,v}]++;
    yal[{v,u}]++;
    if (yal[{u,v}]>1 && u!=v){
        q.push({u,v});
    }
    return ;
}
void mrg(ll u,ll v){
    if (mol[u].size()>mol[v].size()) swap(u,v);
    for (auto t : g[u]){
        add_edge(v,par[t]);
        g[v].pb(par[t]);
    }
    g[u].clear();
    for (auto t : mol[u]){
        par[t]=v;
        mol[v].pb(t);
    }
    mol[u].clear();
    return ;
}
int32_t main(){
    sync;
    ll n;
    cin >> n;
    for (int i=1;i<N;i++) par[i]=i,mol[i].pb(i);
    for (int i=1;i<2*n-1;i++){
        ll u,v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
        add_edge(u,v);
    }
    while(q.size()){
        pii a=q.front();
        q.pop();
       // cout << a.F << " " << a.S << endl;
        mrg(a.F,a.S);
    }
    ll y=par[1];
    for (int i=2;i<=n;i++){
        if (par[i]!=y) kill("NO");
    }
    kill("YES");

}

```