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

Solution (in CPP):

Solution:
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#define fo(i,j,k) for(int i=j;i<=k;++i)
using namespace std;
const int N=1e5+10;
typedef pair<int,int> pr;
typedef multiset<int>::iterator it;

queue<pr> q;
map<pr,int> mp;
multiset<int> b[N];

int f[N];

int find(int x){return !f[x]?x:f[x]=find(f[x]);}

void link(int x,int y){b[x].insert(y),b[y].insert(x);}

pr mkpr(int x,int y){
    if(x>y) swap(x,y);
    return make_pair(x,y);
}

int main() {
    int n;
    scanf("%d",&n);
    fo(i,1,n*2-2){
        int x,y;
        scanf("%d %d",&x,&y);
        link(x,y);
        pr t=mkpr(x,y);
        ++mp[t];
        if(mp[t]==2) q.push(t);
    }
    fo(ch,2,n){
        if(q.empty()) return puts("NO"),0;
        int x=0,y=0;
        for(;x==y;x=find(q.front().first),y=find(q.front().second),q.pop());
        if(b[x].size()>b[y].size()) swap(x,y);
        f[x]=y;
        for(it i=b[x].begin();i!=b[x].end();i++){
            int v=find(*i);
            b[v].erase(b[v].find(x));
            if(v==y) continue;
            link(v,y);
            pr t=mkpr(v,y);
            ++mp[t];
            if(mp[t]==2) q.push(t);
        }
        b[x].clear();
    }
    puts("YES");
}