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<cstdio>
#include<set>
using namespace std;
const int N=100100;
int i,j,k,n,ch,ff,dn,x,y;
int f[N];
set<int> S[N];
set<int>::iterator it;
struct cc { int x,y;} d[N];
void R(int &x) {
	ff=x=0;ch=getchar();
	while (ch<'0' || '9'<ch) ch=getchar();
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
	if (ff) x=-x;
}
int getf(int x) {
	if (f[x]==x) return x;
	return f[x]=getf(f[x]);
}
void Add(int x,int y) {
	x=getf(x);y=getf(y);
	if (S[x].find(y)!=S[x].end()) {
		S[x].erase(y);
		S[y].erase(x);
		d[++dn].x=x;d[dn].y=y;
	}
	else {
		S[x].insert(y);
		S[y].insert(x);
	}
}
void Merge(int x,int y) {
	x=getf(x);y=getf(y);
	if (S[x].size()>S[y].size()) {
		int t=x;x=y;y=t;
	}
	f[x]=y;
	for (it=S[x].begin();it!=S[x].end();it++) {
		int t=*it;
		S[t].erase(x);
		Add(t,y);
	}
}
int main() {
	R(n);
	for (i=1;i<n;i++) f[i]=i;
	for (i=1;i<n;i++) {
		R(x);R(y);
		Add(x,y);
	}
	for (i=1;i<n;i++) {
		R(x);R(y);
		Add(x,y);
	}
	for (i=1;i<n-1;i++) {
		if (i>dn) return puts("NO"),0;
		Merge(d[i].x,d[i].y);
	}
	puts("YES");
}