#include<cstdio>
#include<cctype>
#include<cstdlib>
#include<cstring>
inline int $(){
    register int x = 0;register char ch = getchar();
    while(!isdigit(ch))ch = getchar();
    for(;isdigit(ch);ch = getchar())x = (x << 1) + (x << 3) + (ch ^ 48);
    return x;
}
#define MAXN 100010
#include<set>
#include<map>
#include<queue>
std::set<int> next[MAXN];
std::map<std::pair<int, int>, int> M;
std::queue<std::pair<int, int> > Q;
int n = $();
int p[MAXN];
inline int find(int x){return x == p[x] ? x : p[x] = find(p[x]);}
int main(){
    for(int i = 1;i <= n;i++)p[i] = i;
    for(int i = 2;(i >> 1) < n;i++){
        int x = $(), y = $();
        if(x > y)std::swap(x, y);
        next[x].insert(y);next[y].insert(x);
        if(++M[std::make_pair(x, y)] == 2)Q.push(std::make_pair(x, y));
    }
    for(int i = 1;i < n;i++){
        register int x = 0, y = 0;
        while(x == y){
            if(Q.empty())return puts("NO"), 0;
            x = find(Q.front().first), y = find(Q.front().second);
            Q.pop();
        }
        if(next[x].size() > next[y].size())std::swap(x, y);
        p[x] = y;
        for(auto i : next[x]){
            if(find(i) == y)continue;
            next[y].insert(find(i));
            std::pair<int, int> add = std::make_pair(std::min(find(i), y), std::max(find(i), y));
            if(++M[add] == 2)Q.push(add);
        }
        next[y].erase(x);
    }
    puts("YES");
    return 0;
}
### Prompt

In CPP, your task is to solve 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<cstdio>
#include<cctype>
#include<cstdlib>
#include<cstring>
inline int $(){
    register int x = 0;register char ch = getchar();
    while(!isdigit(ch))ch = getchar();
    for(;isdigit(ch);ch = getchar())x = (x << 1) + (x << 3) + (ch ^ 48);
    return x;
}
#define MAXN 100010
#include<set>
#include<map>
#include<queue>
std::set<int> next[MAXN];
std::map<std::pair<int, int>, int> M;
std::queue<std::pair<int, int> > Q;
int n = $();
int p[MAXN];
inline int find(int x){return x == p[x] ? x : p[x] = find(p[x]);}
int main(){
    for(int i = 1;i <= n;i++)p[i] = i;
    for(int i = 2;(i >> 1) < n;i++){
        int x = $(), y = $();
        if(x > y)std::swap(x, y);
        next[x].insert(y);next[y].insert(x);
        if(++M[std::make_pair(x, y)] == 2)Q.push(std::make_pair(x, y));
    }
    for(int i = 1;i < n;i++){
        register int x = 0, y = 0;
        while(x == y){
            if(Q.empty())return puts("NO"), 0;
            x = find(Q.front().first), y = find(Q.front().second);
            Q.pop();
        }
        if(next[x].size() > next[y].size())std::swap(x, y);
        p[x] = y;
        for(auto i : next[x]){
            if(find(i) == y)continue;
            next[y].insert(find(i));
            std::pair<int, int> add = std::make_pair(std::min(find(i), y), std::max(find(i), y));
            if(++M[add] == 2)Q.push(add);
        }
        next[y].erase(x);
    }
    puts("YES");
    return 0;
}
```