#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pir;

int read() {
    int x = 0, f = 1; char ch = getchar();
    while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
    while (isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

const int Max = 100033;
int n, fa[Max], ans;
set<int> e[Max];
map<pir, int> cnt;
queue<pir> que;
int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int main() {
    n = read();
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 1; i <= 2 * n - 2; i++) {
        int x = read(), y = read();
        e[x].insert(y), e[y].insert(x);
        cnt[pir(x, y)]++, cnt[pir(y, x)]++;
        if (cnt[pir(x, y)] == 2) que.push(pir(x, y));
    }
    while (!que.empty()) {
        int x = que.front().first, y = que.front().second;
        que.pop();
        x = find(x), y = find(y);
        e[x].erase(y), e[y].erase(x);
        if (e[x].size() < e[y].size()) swap(x, y);
        fa[y] = x;
        for (set<int>::iterator i = e[y].begin(); i != e[y].end(); i++) {
            e[*i].erase(y), e[*i].insert(x), e[x].insert(*i);
            cnt[pir(y, *i)] = cnt[pir(*i, y)] = 0;
            cnt[pir(x, *i)]++, cnt[pir(*i, x)]++;
            if (cnt[pir(x, *i)] == 2) que.push(pir(x, *i));
        }
        e[y].clear();
    }
    for (int i = 1; i <= n; i++)
        if (i == find(i)) ans++;
    puts(ans == 1 ? "YES" : "NO");
    return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef pair<int, int> pir;

int read() {
    int x = 0, f = 1; char ch = getchar();
    while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
    while (isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

const int Max = 100033;
int n, fa[Max], ans;
set<int> e[Max];
map<pir, int> cnt;
queue<pir> que;
int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int main() {
    n = read();
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 1; i <= 2 * n - 2; i++) {
        int x = read(), y = read();
        e[x].insert(y), e[y].insert(x);
        cnt[pir(x, y)]++, cnt[pir(y, x)]++;
        if (cnt[pir(x, y)] == 2) que.push(pir(x, y));
    }
    while (!que.empty()) {
        int x = que.front().first, y = que.front().second;
        que.pop();
        x = find(x), y = find(y);
        e[x].erase(y), e[y].erase(x);
        if (e[x].size() < e[y].size()) swap(x, y);
        fa[y] = x;
        for (set<int>::iterator i = e[y].begin(); i != e[y].end(); i++) {
            e[*i].erase(y), e[*i].insert(x), e[x].insert(*i);
            cnt[pir(y, *i)] = cnt[pir(*i, y)] = 0;
            cnt[pir(x, *i)]++, cnt[pir(*i, x)]++;
            if (cnt[pir(x, *i)] == 2) que.push(pir(x, *i));
        }
        e[y].clear();
    }
    for (int i = 1; i <= n; i++)
        if (i == find(i)) ans++;
    puts(ans == 1 ? "YES" : "NO");
    return 0;
}
```