#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

map<pii, int> mp;
set<int> st[100010];
queue<pii> q;
int n, f[100010];

int find(int x)
{
    if (f[x] == x)
        return x;
    return f[x] = find(f[x]);
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n * 2 - 2; ++i)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        st[x].insert(y);
        st[y].insert(x);
        mp[pii(x, y)]++;
        mp[pii(y, x)]++;
        if (mp[pii(x, y)] == 2)
            q.push(pii(x, y));
    }
    for (int i = 1; i <= n; ++i)
        f[i] = i;
    set<int>::iterator it;
    while (!q.empty())
    {
        int x = q.front().first;
        int y = q.front().second;
        q.pop();
        x = find(x);
        y = find(y);
        st[x].erase(y);
        st[y].erase(x);
        if (st[x].size() < st[y].size())
            swap(x, y);
        for (it = st[y].begin(); it != st[y].end(); ++it)
        {
            int z = *it;
            mp[pii(y, z)] = mp[pii(z, y)] = 0;
            st[z].erase(y);
            st[z].insert(x);
            mp[pii(x, z)]++;
            mp[pii(z, x)]++;
            st[x].insert(z);
            if (mp[pii(x, z)] == 2)
                q.push(pii(x, z));
        }
        f[y] = x;
        st[y].clear();
    }
    int s = 0;
    for (int i = 1; i <= n; ++i)
        s += (f[i] == i);
    if (s == 1)
        puts("YES");
    else
        puts("NO");
    return 0;
}
### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

map<pii, int> mp;
set<int> st[100010];
queue<pii> q;
int n, f[100010];

int find(int x)
{
    if (f[x] == x)
        return x;
    return f[x] = find(f[x]);
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n * 2 - 2; ++i)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        st[x].insert(y);
        st[y].insert(x);
        mp[pii(x, y)]++;
        mp[pii(y, x)]++;
        if (mp[pii(x, y)] == 2)
            q.push(pii(x, y));
    }
    for (int i = 1; i <= n; ++i)
        f[i] = i;
    set<int>::iterator it;
    while (!q.empty())
    {
        int x = q.front().first;
        int y = q.front().second;
        q.pop();
        x = find(x);
        y = find(y);
        st[x].erase(y);
        st[y].erase(x);
        if (st[x].size() < st[y].size())
            swap(x, y);
        for (it = st[y].begin(); it != st[y].end(); ++it)
        {
            int z = *it;
            mp[pii(y, z)] = mp[pii(z, y)] = 0;
            st[z].erase(y);
            st[z].insert(x);
            mp[pii(x, z)]++;
            mp[pii(z, x)]++;
            st[x].insert(z);
            if (mp[pii(x, z)] == 2)
                q.push(pii(x, z));
        }
        f[y] = x;
        st[y].clear();
    }
    int s = 0;
    for (int i = 1; i <= n; ++i)
        s += (f[i] == i);
    if (s == 1)
        puts("YES");
    else
        puts("NO");
    return 0;
}
```