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<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
set<int>s[N];
map<pair<int,int>,int> ma;
pair<int,int> q[N];
int n,h,t,fa[N];
int get(int x){
	if(fa[x]==x) return x;
	return fa[x]=get(fa[x]);
}
int main(){
	//freopen("B.in","r",stdin);
	//freopen("B.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n*2-2;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        s[u].insert(v); s[v].insert(u);
        ma[make_pair(u,v)]++; ma[make_pair(v,u)]++;
        if (ma[make_pair(u,v)]==2) q[++t]=make_pair(u,v);
    }
    for (int i=1;i<=n;i++) fa[i]=i;
    set<int>::iterator it;
    while (h<t){
        int u=q[++h].first,v=q[h].second;
        u=get(u); v=get(v);
        s[u].erase(v); s[v].erase(u);
        if (s[u].size()<s[v].size()) swap(u,v);
        for (it=s[v].begin();it!=s[v].end();++it){
            int to=(*it);
            ma[make_pair(to,v)]=ma[make_pair(v,to)]=0;
            s[to].erase(v); s[to].insert(u);
            ma[make_pair(to,u)]++; ma[make_pair(u,to)]++; 
            if (ma[make_pair(to,u)]==2) q[++t]=make_pair(to,u);
            s[u].insert(to);
        }
        fa[v]=u;
        s[v].clear();
    }
    int s=0;
    for (int i=1;i<=n;i++) s+=fa[i]==i;
    if (s==1) printf("YES\n"); else printf("NO\n");
    return 0;
}