#include <bits/stdc++.h>
#define int long long
#define P(x) cout << x << endl
#define D(x) P(#x << ": " << x)
#define F(i,n) for (int i=0; i<(int)(n); i++)
#define DEC(i,n) for (int i=(int)(n); --i>=0;)
#define S(s) (int)((s).size())
#define ALL(v) v.begin(), v.end()
#define MI(a,v) a = min(a,(v))
#define MA(a,v) a = max(a,(v))
#define V vector
#define pb push_back
#define mt make_tuple
using namespace std;
const int N=1e5;

int p[N];
set<int> adj[N];
int find(int a) {return p[a] == a ? a : (p[a] = find(p[a]));}

signed main() {
    int n; cin>>n;
    queue<pair<int,int>> q;
    F(i,2*(n-1)) {
        int a,b; cin>>a>>b;
        a--, b--;
        if (adj[a].count(b))
            q.push({a,b});
        adj[a].insert(b);
        adj[b].insert(a);
    }
    F(i,n) p[i] = i;
    int cnt=0;
    while (!q.empty()) {
        cnt++;
        int a,b;
        tie(a,b) = q.front(); q.pop();
        //D(a), D(b);
        a = find(a), b = find(b);
        if (S(adj[a]) < S(adj[b]))
            swap(a,b);
        p[b] = a;
        for (int v : adj[b]) if (v != a) {
            if (adj[a].count(v))
                q.push({a,v});
            adj[a].insert(v);
            adj[v].erase(b);
            adj[v].insert(a);
        }
    }
    //D(cnt);
    if (cnt == n-1) P("YES");
    else P("NO");
}
### Prompt

Your challenge is to write 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>
#define int long long
#define P(x) cout << x << endl
#define D(x) P(#x << ": " << x)
#define F(i,n) for (int i=0; i<(int)(n); i++)
#define DEC(i,n) for (int i=(int)(n); --i>=0;)
#define S(s) (int)((s).size())
#define ALL(v) v.begin(), v.end()
#define MI(a,v) a = min(a,(v))
#define MA(a,v) a = max(a,(v))
#define V vector
#define pb push_back
#define mt make_tuple
using namespace std;
const int N=1e5;

int p[N];
set<int> adj[N];
int find(int a) {return p[a] == a ? a : (p[a] = find(p[a]));}

signed main() {
    int n; cin>>n;
    queue<pair<int,int>> q;
    F(i,2*(n-1)) {
        int a,b; cin>>a>>b;
        a--, b--;
        if (adj[a].count(b))
            q.push({a,b});
        adj[a].insert(b);
        adj[b].insert(a);
    }
    F(i,n) p[i] = i;
    int cnt=0;
    while (!q.empty()) {
        cnt++;
        int a,b;
        tie(a,b) = q.front(); q.pop();
        //D(a), D(b);
        a = find(a), b = find(b);
        if (S(adj[a]) < S(adj[b]))
            swap(a,b);
        p[b] = a;
        for (int v : adj[b]) if (v != a) {
            if (adj[a].count(v))
                q.push({a,v});
            adj[a].insert(v);
            adj[v].erase(b);
            adj[v].insert(a);
        }
    }
    //D(cnt);
    if (cnt == n-1) P("YES");
    else P("NO");
}
```