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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class TEST{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		
		
		int[] depth = new int[N];
		int[] parent = new int[N];
		var Q = new Queue<int>();
		Q.Enqueue(0);
		parent[0] = -1;
		while(Q.Count > 0){
			var now = Q.Dequeue();
			foreach(var nxt in E[now]){
				if(nxt == parent[now]) continue;
				parent[nxt] = now;
				depth[nxt] = depth[now] + 1;
				Q.Enqueue(nxt);
			}
		}
		
		Pair[] P = new Pair[N];
		for(int i=0;i<N;i++) P[i] = new Pair(i, depth[i]);
		Array.Sort(P, (p,q) => p.Depth.CompareTo(q.Depth));
		Array.Reverse(P);
		
		long[] sum = new long[N];
		for(int i=1;i<N;i++){
			if(E[i].Count == 1) sum[i] = A[i];
		}
		
		
		foreach(var p in P){
			var now = p.Idx;
			if(now == 0) break;
			if(E[now].Count == 1){
				sum[parent[now]] += A[now];
				continue;
			}
			if(E[now].Count == 2){
				if(A[now] != sum[now]){
					Console.WriteLine("NO");
					return;
				}
				sum[parent[now]] += A[now];
				continue;
			}
			
			long v = sum[now] - A[now];
			if(v < 0){
				Console.WriteLine("NO");
				return;
			}
			long up = sum[now] - 2 * v;
			if(up < 0){
				Console.WriteLine("NO");
				return;
			}
			sum[parent[now]] += up;
		}
		
		if(E[0].Count == 1){
			Console.WriteLine(sum[0] == A[0] ? "YES" : "NO");
		} else {
			Console.WriteLine(sum[0] - 2 * (sum[0] - A[0]) == 0? "YES" : "NO");
		}
		
		
	}
	
	class Pair{
		public int Idx,Depth;
		public Pair(int idx, int depth){
			Idx = idx; Depth = depth;
		}
	}
	
	int N;
	long[] A;
	List<int>[] E;
	public Sol(){
		N = ri();
		A = rla();
		E = new List<int>[N];
		for(int i=0;i<N;i++) E[i] = new List<int>();
		for(int i=0;i<N-1;i++){
			var d = ria();
			E[d[0] - 1].Add(d[1] - 1);
			E[d[1] - 1].Add(d[0] - 1);
		}
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
}