//☭
#include<cstdio>
#include<queue>
#include<map>
#include<set>
 
using namespace std;
 
const int N=1e5+10;
 
struct pai{
    int f, s;
};
 
bool operator <(pai a, pai b) {
    return (a.f==b.f)?(a.s<b.s):(a.f<b.f);
}
 
map< pai, int>m;
set<int> mu[N];
queue< pai >q;
 
int f[N], n;
 
inline void add(int x, int y) {mu[x].insert(y), mu[y].insert(x);}
 
pai marry(int x, int y) {
    if(x>y) {int t=y; y=x, x=t;}
    return (pai){x, y};
}
 
int f_f(int x){
    return f[x]?f[x]=f_f(f[x]):x;
}
 
int main(){
    scanf("%d", &n);
    for(int i=1; i<=(n<<1)-2; i++) {
        int x, y; scanf("%d%d", &x, &y), add(x, y);
        pai edg=marry(x, y); ++m[edg];
        if(m[edg]==2) q.push(edg);
        if(m[edg]>2) {puts("NO"); return 0;}
    }
    for(int w=1; w<n; w++) {
        if(q.empty()) {puts("NO"); return 0;}
        int x=0, y=0;
        while(x==y) x=f_f(q.front().f), y=f_f(q.front().s), q.pop();
        if(mu[x].size()>mu[y].size()) swap(x, y); f[x]=y;
        for(set<int>::iterator i=mu[x].begin(); i!=mu[x].end(); i++){
            int from=f_f(*i); mu[from].erase(mu[from].find(x)); if(from==y) continue;
            add(from,y); pai t=marry(from,y); m[t]++; if(m[t]==2) q.push(t);
        }
        mu[x].clear();
    }
    puts("YES");
}

### Prompt

Generate a Cpp solution to 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<queue>
#include<map>
#include<set>
 
using namespace std;
 
const int N=1e5+10;
 
struct pai{
    int f, s;
};
 
bool operator <(pai a, pai b) {
    return (a.f==b.f)?(a.s<b.s):(a.f<b.f);
}
 
map< pai, int>m;
set<int> mu[N];
queue< pai >q;
 
int f[N], n;
 
inline void add(int x, int y) {mu[x].insert(y), mu[y].insert(x);}
 
pai marry(int x, int y) {
    if(x>y) {int t=y; y=x, x=t;}
    return (pai){x, y};
}
 
int f_f(int x){
    return f[x]?f[x]=f_f(f[x]):x;
}
 
int main(){
    scanf("%d", &n);
    for(int i=1; i<=(n<<1)-2; i++) {
        int x, y; scanf("%d%d", &x, &y), add(x, y);
        pai edg=marry(x, y); ++m[edg];
        if(m[edg]==2) q.push(edg);
        if(m[edg]>2) {puts("NO"); return 0;}
    }
    for(int w=1; w<n; w++) {
        if(q.empty()) {puts("NO"); return 0;}
        int x=0, y=0;
        while(x==y) x=f_f(q.front().f), y=f_f(q.front().s), q.pop();
        if(mu[x].size()>mu[y].size()) swap(x, y); f[x]=y;
        for(set<int>::iterator i=mu[x].begin(); i!=mu[x].end(); i++){
            int from=f_f(*i); mu[from].erase(mu[from].find(x)); if(from==y) continue;
            add(from,y); pai t=marry(from,y); m[t]++; if(m[t]==2) q.push(t);
        }
        mu[x].clear();
    }
    puts("YES");
}

```