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;

int main()
{
    int N; cin >> N;

    auto adjacency_list = vector<map<int, int>>(N);
    auto queue = std::queue<pair<int, int>>();
    for (auto i = 0; i < 2*(N-1); i++)
    {
        int a, b; cin >> a >> b;
        a--; b--;

        adjacency_list[a][b]++;
        adjacency_list[b][a]++;

        if (adjacency_list[a][b] == 2)
            queue.push({a, b});
    }

    auto representatives = vector<int>(N);
    iota(begin(representatives), end(representatives), 0);

    const function<int(int)> find = [&](const int i)
    {
        if (representatives[i] != i)
            representatives[i] = find(representatives[i]);
        return representatives[i];
    };

    auto size = N;
    while (!queue.empty())
    {
        int a, b; tie(a, b) = queue.front(); queue.pop();
        a = find(a); b = find(b);

        if (adjacency_list[a].size() > adjacency_list[b].size())
            swap(a, b);

        representatives[a] = b;

        adjacency_list[a].erase(b);
        adjacency_list[b].erase(a);

        for (const auto i : adjacency_list[a])
        {
            const auto v = i.first, x = i.second;
            adjacency_list[v].erase(a);
            adjacency_list[v][b] += x;
            adjacency_list[b][v] += x;
            if (x != 2 && adjacency_list[b][v] == 2)
                queue.push({b, v});
        }

        size--;
    }

    cout << (size == 1 ? "YES" : "NO") << endl;
}