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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

struct UnionFind
{
  vector< int > data;

  UnionFind(int sz)
  {
    data.assign(sz, -1);
  }

  bool unite(int x, int y)
  {
    x = find(x), y = find(y);
    if(x == y) return (false);
    data[x] += data[y];
    data[y] = x;
    return (true);
  }

  int find(int k)
  {
    if(data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }

  int size(int k)
  {
    return (-data[find(k)]);
  }
};

int main()
{
  int N;
  set< int > g[100000];
  cin >> N;
  queue< pair< int, int > > que;
  for(int i = 0; i < 2 * N - 2; i++) {
    int A, B;
    cin >> A >> B;
    --A, --B;
    if(g[A].count(B)) que.emplace(A, B);
    g[A].emplace(B);
    g[B].emplace(A);
  }
  UnionFind tree(N);
  while(!que.empty()) {
    int a, b;
    tie(a, b) = que.front();
    que.pop();
    a = tree.find(a), b = tree.find(b);
    if(a == b) continue;
    if(g[a].size() > g[b].size()) swap(a, b);
    tree.unite(b, a);
    g[a].erase(b);
    g[b].erase(a);
    for(int to : g[a]) {
      g[to].erase(a);
      if(g[to].count(b)) que.emplace(to, b);
      else g[to].emplace(b), g[b].emplace(to);
    }
    g[a].clear();
  }
  if(tree.size(0) == N) cout << "YES" << endl;
  else cout << "NO" << endl;
}