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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define tree_valtype int

typedef struct tree_edge_sub tree_edge;

typedef struct {
	int num;
	int nearnum;
	tree_edge *near;
}tree_vertex_sub;

struct tree_edge_sub{
	tree_vertex_sub *v;
	int w;
	tree_edge *next;
};

typedef struct tree_v_sub tree_vertex;

struct tree_v_sub{
	int num;
	tree_valtype val;
	tree_vertex *parent;
	int pareweight;
	int chilnum;
	tree_vertex **children;
	int *chilweight;
};

typedef struct {
	int N;
	int root;
	tree_vertex_sub **v_s;
	tree_vertex **v;
	tree_vertex **sorted_v;
}tree;

//頂点数N, 根の番号root, 各頂点の初期値ini_valの木を作る
tree *make_tree(int N, int root, tree_valtype ini_val){
	int i;
	tree *t = (tree *)malloc(sizeof(tree));
	t->N = N;
	t->root = root;
	t->v_s = (tree_vertex_sub **)malloc(sizeof(tree_vertex_sub *) * N);
	t->v = (tree_vertex **)malloc(sizeof(tree_vertex *) * N);
	t->sorted_v = (tree_vertex **)malloc(sizeof(tree_vertex *) * N);
	tree_vertex *parent_in_law = (tree_vertex *)malloc(sizeof(tree_vertex));
	parent_in_law->num = -1;
	parent_in_law->val = ini_val;
	parent_in_law->parent = NULL;
	parent_in_law->pareweight = -1;
	parent_in_law->chilnum = 0;
	parent_in_law->children = NULL;
	parent_in_law->chilweight = NULL;
	for(i = 0; i < N; i++){
		(t->v_s)[i] = (tree_vertex_sub *)malloc(sizeof(tree_vertex_sub));
		(t->v_s)[i]->num = i;
		(t->v_s)[i]->nearnum = 0;
		(t->v_s)[i]->near = NULL;
		(t->v)[i] = (tree_vertex *)malloc(sizeof(tree_vertex));
		(t->v)[i]->num = i;
		(t->v)[i]->val = ini_val;
		(t->v)[i]->parent = parent_in_law;
		(t->v)[i]->pareweight = -1;
		(t->v)[i]->chilnum = 0;
		(t->v)[i]->children = NULL;
		(t->v)[i]->chilweight = NULL;
		(t->sorted_v)[i] = NULL;
	}
	return t;
}

//木tの頂点aと頂点bの間に重みwの無向辺を張る (0 <= a, b <= N - 1)
void set_edge_tree(int a, int b, int w, tree *t){
	tree_edge *new1 = (tree_edge *)malloc(sizeof(tree_edge));
	new1->v = (t->v_s)[b];
	new1->w = w;
	new1->next = (t->v_s)[a]->near;
	(t->v_s)[a]->near = new1;
	(t->v_s)[a]->nearnum++;

	tree_edge *new2 = (tree_edge *)malloc(sizeof(tree_edge));
	new2->v = (t->v_s)[a];
	new2->w = w;
	new2->next = (t->v_s)[b]->near;
	(t->v_s)[b]->near = new2;
	(t->v_s)[b]->nearnum++;
}

//set_edge_tree後に呼び出す
void build_tree(tree *t){
	int i, j;
	tree_vertex_sub **v_s = t->v_s;
	tree_vertex **v = t->v;
	tree_vertex **sorted_v = t->sorted_v;
	sorted_v[0] = v[t->root];
	tree_vertex *nowv;
	tree_edge *nowe;
	for(i = 0, j = 1; j - i > 0; i++){
		nowv = sorted_v[i];
		if(i == 0){
			v_s[nowv->num]->nearnum++;
		}
		nowv->children = (tree_vertex **)malloc(sizeof(tree_vertex *) * (v_s[nowv->num]->nearnum - 1));
		nowv->chilweight = (int *)malloc(sizeof(int) * (v_s[nowv->num]->nearnum - 1));
		if(i == 0){
			v_s[nowv->num]->nearnum--;
		}
		for(nowe = v_s[nowv->num]->near; nowe != NULL; nowe = nowe->next){
			if(nowe->v->num != nowv->parent->num){
				(nowv->children)[nowv->chilnum] = v[nowe->v->num];
				(nowv->chilweight)[nowv->chilnum] = nowe->w;
				nowv->chilnum++;
				v[nowe->v->num]->parent = nowv;
				v[nowe->v->num]->pareweight = nowe->w;
				sorted_v[j] = v[nowe->v->num];
				j++;
			}
		}
	}
}

int main(){
	int N, a, b, i, nowA, nowS;
	scanf("%d", &N);
	int *A = (int *)malloc(sizeof(int) * N);
	int *S = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d", &A[i]);
		S[i] = 0;
	}
	tree *t = make_tree(N, 0, 0);
	for(i = 1; i < N; i++){
		scanf("%d%d", &a, &b);
		set_edge_tree(a - 1, b - 1, 1, t);
	}
	if(N == 2){
		if(A[0] == A[1]){
			printf("YES\n");
		}
		else{
			printf("NO\n");
		}
		return 0;
	}
	for(i = 0; i < N; i++){
		if(t->v_s[i]->nearnum > 1){
			t->root = i;
			break;
		}
	}
	build_tree(t);
	tree_vertex *nowv;
	for(i = N - 1; i > 0; i--){
//		printf("i = %d\n", i);
		nowv = t->sorted_v[i];
//		printf("nowv->num = %d\n", nowv->num);
		nowA = A[nowv->num];
		nowS = S[nowv->num];
		if(nowv->chilnum == 0){
			if(A[nowv->parent->num] < nowA){
				printf("NO\n");
				return 0;
			}
			S[nowv->parent->num] += nowA;
		}
		else{
			if(nowA <= nowS && nowS <= 2 * nowA){
				if(A[nowv->parent->num] < 2 * nowA - nowS){
					printf("NO\n");
					return 0;
				}
				S[nowv->parent->num] += 2 * nowA - nowS;
			}
			else{
				printf("NO\n");
				return 0;
			}
		}
	}
	nowv = t->sorted_v[0];
	nowA = A[nowv->num];
	nowS = S[nowv->num];
	if(2 * nowA == nowS){
		printf("YES\n");
	}
	else{
		printf("NO\n");
	}
	return 0;
}