图论水题但是没想到

Note that this is the first problem of the two similar problems. You can hack this problem only if you solve both problems.

You are given a tree with nn nodes. In the beginning, 00 is written on all edges. In one operation, you can choose any 22 distinct leaves uu, vv and any real number xx and add xx to values written on all edges on the simple path between uu and vv.

For example, on the picture below you can see the result of applying two operations to the graph: adding 22 on the path from 77 to 66, and then adding −0.5−0.5 on the path from 44 to 55.

Is it true that for any configuration of real numbers written on edges, we can achieve it with a finite number of operations?

Leaf is a node of a tree of degree 11. Simple path is a path that doesn't contain any node twice.

Input

The first line contains a single integer nn (2≤n≤1052≤n≤105) — the number of nodes.

Each of the next n−1n−1 lines contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v), meaning that there is an edge between nodes uu and vv. It is guaranteed that these edges form a tree.

Output

If there is a configuration of real numbers written on edges of the tree that we can't achieve by performing the operations, output "NO".

Otherwise, output "YES".

You can print each letter in any case (upper or lower).

Examples

Input

2
1 2

Output

YES

Input

3
1 2
2 3

Output

NO

Input

5
1 2
1 3
1 4
2 5

Output

NO

Input

6
1 2
1 3
1 4
2 5
2 6

Output

YES

Note

In the first example, we can add any real xx to the value written on the only edge (1,2)(1,2).

In the second example, one of configurations that we can't reach is 00 written on (1,2)(1,2) and 11 written on (2,3)(2,3).

Below you can see graphs from examples 33, 44:

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;

ll in[200001]; 

int main(){
	ll n;
	while(cin>>n){
		memset(in,0,sizeof(in));
		for(ll i=1;i<=n-1;i++){
			ll x,y;
			cin>>x>>y;
			in[x]++;
			in[y]++;
		}
		ll sign=0;
		for(ll i=1;i<=n;i++){
			if(in[i]==2){
				sign=1;
				break;
			}
		}
		if(sign==1){
			cout<<"NO"<<endl;
		}
		else cout<<"YES"<<endl;
	}
	return 0;
}

　　

 

原文链接:https://www.cnblogs.com/akpower/p/11284887.html
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