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;
typedef long long ll;
typedef pair<int, int> pii;

const int MAXN = 1e5+5;

int n;
set<int> a[MAXN][2];
map<pii, int> m;
set<pii> s;

void rem(int u, int v, int x) {
    if (u > v)
        swap(u, v);
    m[pii(u, v)]--;
    if (m[pii(u, v)] < 2)
        s.erase(pii(u, v));
    a[u][x].erase(v);
    a[v][x].erase(u);
}

void add(int u, int v, int x) {
    if (u > v)
        swap(u, v);
    m[pii(u, v)]++;
    if (m[pii(u, v)] >= 2)
        s.insert(pii(u, v));
    a[u][x].insert(v);
    a[v][x].insert(u);
}

void merge(int u, int v) {
    rem(u, v, 0);
    rem(u, v, 1);

    if (a[u][0].size()+a[u][1].size() > a[v][0].size()+a[v][1].size())
        swap(u, v);
    set<int> s0 = a[u][0];
    set<int> s1 = a[u][1];

    for (int w : s0) {
        rem(u, w, 0);
        add(v, w, 0);
    }

    for (int w : s1) {
        rem(u, w, 1);
        add(v, w, 1);
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n-1; ++i) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        add(u, v, 0);
    }

    for (int i = 0; i < n-1; ++i) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        add(u, v, 1);
    }

    for (int i = 0; i < n-1; ++i) {
        if (s.empty()) {
            cout << "NO\n";
            return 0;
        }

        pii t = (*s.begin());
        merge(t.first, t.second);
    }

    cout << "YES\n";
}