#include <bits/stdc++.h>
using namespace std;

const int N = 100100;

int n;
set<int> g[N];
int p[N];

int get(int x) {
    if (p[x] == -1) {
        return x;
    }
    return p[x] = get(p[x]);
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a--; b--;
        g[a].insert(b);
        g[b].insert(a);
    }

    vector<pair<int, int>> cand;   
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a--; b--;
        if (g[a].count(b)) {
            cand.emplace_back(a, b);
        }
        g[a].insert(b);
        g[b].insert(a);
    }

    fill_n(p, n, -1);
    int cnt = 0;
    while (!cand.empty()) {
        int a, b;
        tie(a, b) = cand.back();
        cand.pop_back();
        a = get(a);
        b = get(b);
        if (a == b) {
            continue;
        }

        cnt++;

        if (g[a].size() > g[b].size()) {
            swap(a, b);
        }

        p[a] = b;

        g[a].erase(b);
        g[b].erase(a);
        for (int x: g[a]) {
            g[x].erase(a);
            if (g[x].count(b)) {
                cand.emplace_back(x, b);
            }
            g[x].insert(b);
            g[b].insert(x);
        }
        set<int>().swap(g[a]);
    }
    
    if (cnt == n - 1) {
        puts("YES");
    } else {
        puts("NO");
    }
}

### 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;

const int N = 100100;

int n;
set<int> g[N];
int p[N];

int get(int x) {
    if (p[x] == -1) {
        return x;
    }
    return p[x] = get(p[x]);
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a--; b--;
        g[a].insert(b);
        g[b].insert(a);
    }

    vector<pair<int, int>> cand;   
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        a--; b--;
        if (g[a].count(b)) {
            cand.emplace_back(a, b);
        }
        g[a].insert(b);
        g[b].insert(a);
    }

    fill_n(p, n, -1);
    int cnt = 0;
    while (!cand.empty()) {
        int a, b;
        tie(a, b) = cand.back();
        cand.pop_back();
        a = get(a);
        b = get(b);
        if (a == b) {
            continue;
        }

        cnt++;

        if (g[a].size() > g[b].size()) {
            swap(a, b);
        }

        p[a] = b;

        g[a].erase(b);
        g[b].erase(a);
        for (int x: g[a]) {
            g[x].erase(a);
            if (g[x].count(b)) {
                cand.emplace_back(x, b);
            }
            g[x].insert(b);
            g[b].insert(x);
        }
        set<int>().swap(g[a]);
    }
    
    if (cnt == n - 1) {
        puts("YES");
    } else {
        puts("NO");
    }
}

```