Problem:
Score : 700 points

Problem Statement

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.

Currently, there are A_i stones placed on vertex i.
Determine whether it is possible to remove all the stones from the vertices by repeatedly performing the following operation:

Select a pair of different leaves. Then, remove exactly one stone from every vertex on the path between those two vertices.
Here, a leaf is a vertex of the tree whose degree is 1, and the selected leaves themselves are also considered as vertices on the path connecting them.

Note that the operation cannot be performed if there is a vertex with no stone on the path.

Constraints

2 ≦ N ≦ 10^5

1 ≦ a_i,b_i ≦ N

0 ≦ A_i ≦ 10^9

The given graph is a tree.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

If it is possible to remove all the stones from the vertices, print YES. Otherwise, print NO.

Sample Input 1

5
1 2 1 1 2
2 4
5 2
3 2
1 3

Sample Output 1

YES

All the stones can be removed, as follows:

Select vertices 4 and 5. Then, there is one stone remaining on each vertex except 4.

Select vertices 1 and 5. Then, there is no stone on any vertex.

Sample Input 2

3
1 2 1
1 2
2 3

Sample Output 2

NO

Sample Input 3

6
3 2 2 2 2 2
1 2
2 3
1 4
1 5
4 6

Sample Output 3

YES

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define go(i,x,a) for(int i=a;i<x;i++)
#define ll long long
using namespace std;
const int maxn=2e5+5;
int n,m,a[maxn],head[maxn],k,in[maxn],rt;
int tl;
bool ans=0;
ll up[maxn],q[maxn];

struct edd{
	int nxt,v;
}e[maxn*2];

inline void add(int a,int b){
	e[k].v=b; e[k].nxt=head[a]; head[a]=k++;
	e[k].v=a; e[k].nxt=head[b]; head[b]=k++;
}

inline int rd(){
	int ret=0,af=1; char gc=getchar();
	while(gc < '0' || gc > '9'){ if(gc=='-') af=-af; gc=getchar(); }
	while(gc >= '0' && gc <= '9') ret=ret*10+gc-'0',gc=getchar();
	return ret*af;
}

void dfs(int u,int fa){
	if(ans) return;
	ll f,sf=0,sum,d;
	tl=0;
	for(int i=head[u];i+1;i=e[i].nxt){
		int v=e[i].v;
		if(v==fa) continue;
		dfs(v,u);
		if(ans) return;
		q[++tl]=up[v],sum+=up[v];
	}
	if(!tl) up[u]=a[u];
	else{
		f=2*a[u]-sum;d=sum-f;
		if(f<0){ans=1;return;}
		for(int i=1;i<=tl;++i){
			sf+=q[i]>d/2?q[i]-d/2:0;
		}
		if(sf>f){ans=1;return;}
		//我们在“子节点自行解决的时候 ”，子节点总权值-2，父节点权值-1，已排除了一些情况 
		//然而子节点之间的差值过大的时候，同样是不可行的
		//对于一个较大的子节点，如果他的权值数大于“自行解决”的总次数，就必须通过父节点解决，将这些“必须通过父节点解决的权值”加起来与“应该通过父节点解决的权值”比较
		//若sf<f，说明权值比较均匀
		up[u]=f;
	}
}

int main(){
	//freopen("input.txt","r",stdin);
	memset(head,-1,sizeof(head)); k=0;
	n=rd(); int x,y;
	go(i,n+1,1) a[i]=rd();
	go(i,n,1){ x=rd(); y=rd(); add(x,y); in[x]++; in[y]++; }
	if(n==2){
		if(a[1]!=a[2]) printf("NO");
		else puts("YES");
	}
	else{
		go(i,1,n){
			if(in[i]>1){
				x=i;break;
			}
		}
		dfs(x,0);
		if(ans) printf("YES");
		else puts("NO");
	}
}