#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int n, pa[100001];
map<int, int> tr[100001];

int P(int v){
    if (pa[v])return pa[v] = P(pa[v]);
    return v;
}

int main(){
    scanf("%d", &n);
    f(i, 2, n << 1){
        int a, b;
        scanf("%d%d", &a, &b);
        ++tr[a][b];
        ++tr[b][a];
    }
    stack<pair<int, int> > st;
    f(i, 1, n + 1)for (map<int, int>::iterator it = tr[i].lower_bound(i); it != tr[i].end(); ++it)if (it->second >= 2)st.push(make_pair(i, it->first));
    while (!st.empty()){
        int a = P(st.top().first), b = P(st.top().second);
        st.pop();
        if (a == b)continue;
        if (tr[a].size() > tr[b].size())swap(a, b);
        for (map<int, int>::iterator it = tr[a].begin(); it != tr[a].end(); tr[a].erase(it++)){
            if (it->first == b){
                tr[b].erase(a);
                continue;
            }
            map<int, int>::iterator it0 = tr[b].insert(make_pair(it->first, 0)).first;
            int t = it0->second;
            it0->second += it->second;
            tr[it->first][b] += it->second;
            if (t < 2 && it0->second >= 2)st.push(make_pair(it->first, b));
            tr[it->first].erase(a);
        }
        pa[a] = b;
        --n;
    }
    if (n == 1)printf("YES\n");
    else printf("NO\n");
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int n, pa[100001];
map<int, int> tr[100001];

int P(int v){
    if (pa[v])return pa[v] = P(pa[v]);
    return v;
}

int main(){
    scanf("%d", &n);
    f(i, 2, n << 1){
        int a, b;
        scanf("%d%d", &a, &b);
        ++tr[a][b];
        ++tr[b][a];
    }
    stack<pair<int, int> > st;
    f(i, 1, n + 1)for (map<int, int>::iterator it = tr[i].lower_bound(i); it != tr[i].end(); ++it)if (it->second >= 2)st.push(make_pair(i, it->first));
    while (!st.empty()){
        int a = P(st.top().first), b = P(st.top().second);
        st.pop();
        if (a == b)continue;
        if (tr[a].size() > tr[b].size())swap(a, b);
        for (map<int, int>::iterator it = tr[a].begin(); it != tr[a].end(); tr[a].erase(it++)){
            if (it->first == b){
                tr[b].erase(a);
                continue;
            }
            map<int, int>::iterator it0 = tr[b].insert(make_pair(it->first, 0)).first;
            int t = it0->second;
            it0->second += it->second;
            tr[it->first][b] += it->second;
            if (t < 2 && it0->second >= 2)st.push(make_pair(it->first, b));
            tr[it->first].erase(a);
        }
        pa[a] = b;
        --n;
    }
    if (n == 1)printf("YES\n");
    else printf("NO\n");
}
```