#include <bits/stdc++.h>

using namespace std;

//#define FILE_IO

typedef pair<int, int> pii;

int N, f[100005], sz[100005];
queue <pii> q;
multiset <int> edg[100005];

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

int main()
{
    #ifdef FILE_IO
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    #endif

    scanf("%d", &N);
    for(int i = 1; i <= 2 * N - 2; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);

        edg[x].insert(y);
        edg[y].insert(x);

        if(edg[x].count(y) == 2)    q.push({x, y});
    }

    for(int i = 1; i <= N; i++)
        f[i] = i, sz[i] = 1;

    while(!q.empty())
    {
        int x = q.front().first;
        int y = q.front().second;
        q.pop();

        int fx = F(x), fy = F(y);
        if(fx == fy)    continue;
        if(fx != x || fy != y)  continue;

        if(sz[x] > sz[y]) swap(x, y);

        f[x] = y;
        sz[y] += sz[x];

        edg[y].erase(x);
        edg[x].erase(y);

        for(auto nxt: edg[x])
        {
            edg[nxt].erase(x);

            edg[y].insert(nxt);
            edg[nxt].insert(y);

            if(edg[y].count(nxt) == 2)
                q.push({y, nxt});
        }
    }

    if(sz[F(1)] == N)
        printf("YES\n");
    else
        printf("NO\n");

    return 0;
}

### Prompt

Your challenge is to write 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;

//#define FILE_IO

typedef pair<int, int> pii;

int N, f[100005], sz[100005];
queue <pii> q;
multiset <int> edg[100005];

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

int main()
{
    #ifdef FILE_IO
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    #endif

    scanf("%d", &N);
    for(int i = 1; i <= 2 * N - 2; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);

        edg[x].insert(y);
        edg[y].insert(x);

        if(edg[x].count(y) == 2)    q.push({x, y});
    }

    for(int i = 1; i <= N; i++)
        f[i] = i, sz[i] = 1;

    while(!q.empty())
    {
        int x = q.front().first;
        int y = q.front().second;
        q.pop();

        int fx = F(x), fy = F(y);
        if(fx == fy)    continue;
        if(fx != x || fy != y)  continue;

        if(sz[x] > sz[y]) swap(x, y);

        f[x] = y;
        sz[y] += sz[x];

        edg[y].erase(x);
        edg[x].erase(y);

        for(auto nxt: edg[x])
        {
            edg[nxt].erase(x);

            edg[y].insert(nxt);
            edg[nxt].insert(y);

            if(edg[y].count(nxt) == 2)
                q.push({y, nxt});
        }
    }

    if(sz[F(1)] == N)
        printf("YES\n");
    else
        printf("NO\n");

    return 0;
}

```