#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;
const int N = 100005;
set<int> g[N];
int p[N];
int num = 0;
int find(int x){
    return x == p[x] ? x : p[x]=find(p[x]);
}
void merge(int u, int v){
    u = find(u), v = find(v);
    if (u == v) return;
    if (g[u].size() < g[v].size()) swap(u,v);
    //printf("merge %d to %d\n",v,u);
    set<ii> que;
    for (auto &x: g[v]){
        if (x == u) continue;
        //printf("%d connected to %d\n",x,v);
        g[x].erase(v);
        if (g[x].count(u)){
            que.insert({x,u});
        }
        g[u].insert(x);
        g[x].insert(u);
    }
    g[u].erase(v);
    g[v].clear();
    p[v] = u;
    for (auto cur : que){
        merge(cur.first,cur.second);
    }
}
int main(){
    int n;
    scanf("%d",&n);
    iota(p,p+N,0);
    for (int i = 1; i < n; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        g[a].insert(b);
        g[b].insert(a);

    }
    for (int i = 1; i < n; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        a = find(a); b = find(b);
        if (g[a].count(b)) merge(a,b);
        else{
            g[a].insert(b);
            g[b].insert(a);
        }
    }
    int c = find(1);
    for (int i = 2; i <= n; i++){
        if (c != find(i)){
            printf("NO");
            return 0;
        }
    }
    printf("YES");
}

### Prompt

Please provide a Cpp coded solution 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> ii;
const int N = 100005;
set<int> g[N];
int p[N];
int num = 0;
int find(int x){
    return x == p[x] ? x : p[x]=find(p[x]);
}
void merge(int u, int v){
    u = find(u), v = find(v);
    if (u == v) return;
    if (g[u].size() < g[v].size()) swap(u,v);
    //printf("merge %d to %d\n",v,u);
    set<ii> que;
    for (auto &x: g[v]){
        if (x == u) continue;
        //printf("%d connected to %d\n",x,v);
        g[x].erase(v);
        if (g[x].count(u)){
            que.insert({x,u});
        }
        g[u].insert(x);
        g[x].insert(u);
    }
    g[u].erase(v);
    g[v].clear();
    p[v] = u;
    for (auto cur : que){
        merge(cur.first,cur.second);
    }
}
int main(){
    int n;
    scanf("%d",&n);
    iota(p,p+N,0);
    for (int i = 1; i < n; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        g[a].insert(b);
        g[b].insert(a);

    }
    for (int i = 1; i < n; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        a = find(a); b = find(b);
        if (g[a].count(b)) merge(a,b);
        else{
            g[a].insert(b);
            g[b].insert(a);
        }
    }
    int c = find(1);
    for (int i = 2; i <= n; i++){
        if (c != find(i)){
            printf("NO");
            return 0;
        }
    }
    printf("YES");
}

```