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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
                ____________    ______________       __
               / _________  /\ /_____   _____/\     / /\
              / /\       / /  \\    /  /\    \ \   / /  \
             / /  \_____/ /   / \__/  /  \____\/  / /   /
            / /   /    / /   /    /  /   /       / /   /
           / /   /    / /   /    /  /   /       / /   /
          / /   /    / /   /    /  /   /       / /   /
         / /___/____/ /   /    /  /   /       / /___/________
        /____________/   /    /__/   /       /______________/\
        \            \  /     \  \  /        \              \ \
         \____________\/       \__\/          \______________\/
           ___       ___               ___    __________
          /  /\     /  /\             /  /\  /_______  /\
         /  /__\___/  /  \           /  /  \ \      /  /  \
        /____    ____/   /          /  /   /  \____/  /   /
        \   /   /\   \  /          /  /   /       /  /   /
         \_/   /  \___\/ ___      /  /   /       /  /   /
          /   /   /     /  /\    /  /   /       /  /   /
         /   /   /     /  /__\__/  /   /       /  /___/____
        /___/   /     /___________/   /       /___________/\
        \   \  /      \           \  /        \           \ \
         \___\/        \___________\/          \___________\/
 
          A FAN OF FIZZYDAVID
 
*/
 
#include<bits/stdc++.h>
 
#define HEAP priority_queue
#define rep(i,n) for(int i=0;i<(n);i++)
#define per(i,n) for(int i=(n)-1;i>=0;i--)
#define forn(i,l,r) for(int i=(l);i<=(r);i++)
#define nrof(i,r,l) for(int i=(r);i>=(l);i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define X first
#define Y second
#define eps 1e-6
#define pi 3.1415926535897932384626433832795
#define orz int
#define yjz main
#define fizzydavid return
#define ak 0
#define la ;
#define SZ(x) (int)x.size()
#define ALL(x) x.begin(),x.end()
#define FILL(a,b) memset((a),(b),sizeof((a)))
 
using namespace std;
 
typedef long long LL;
typedef double flt;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<LL> vl;
typedef pair<int,LL> pil;
typedef pair<LL,int> pli;
typedef pair<LL,LL> pll;
typedef vector<pil> vil;
typedef vector<pii> vii;
 
const int iinf=1e9+7;
const LL linf=1ll<<60;
const flt dinf=1e10;
 
template <typename T>
inline void scf(T &x)
{
	bool f=0; x=0; char c=getchar();
	while((c<'0' || c>'9') && c!='-') c=getchar();
	if(c=='-') { f=1; c=getchar(); }
	while(c>='0' && c<='9') { x=x*10+c-'0'; c=getchar(); }
	if(f) x=-x; return;
}
 
template <typename T1,typename T2>
void scf(T1 &x,T2 &y) { scf(x); return scf(y); }
 
template <typename T1,typename T2,typename T3>
void scf(T1 &x,T2 &y,T3 &z) { scf(x); scf(y); return scf(z); }
 
template <typename T1,typename T2,typename T3,typename T4>
void scf(T1 &x,T2 &y,T3 &z,T4 &w) { scf(x); scf(y); scf(z); return scf(w); }
 
#ifdef ONLINE_JUDGE
#define debug(x,c) ;
#else
#define DEBUG
#define debug(x,c) cerr<<#x<<"="<<x<<c;
#endif
 
//---------------------------head----------------------------
 
const int N = 1e5 + 100;
 
int n;
vi g[N];
int a[N];
 
void dfs(int u, int fa)
{
    if(SZ(g[u]) == 1) return;
    int tot, Max; tot = Max = 0;
    for(int v: g[u]) if(v != fa) dfs(v, u), tot += a[v], Max = max(Max, a[v]);
    if(tot < a[u]){ puts("NO"); exit(0); }
    if(Max > tot - Max && Max > a[u]){ puts("NO"); exit(0); }
    a[u] = tot - (tot - a[u]) * 2;
    if(a[u] < 0){ puts("NO"); exit(0); }
    return;
}
 
orz yjz()
{
    scf(n);
    forn(i, 1 ,n) scf(a[i]);
    rep(i, n - 1)
    {
        int u, v; scf(u, v);
        g[u].pb(v); g[v].pb(u);
    }
    if(n == 2){ puts(a[1] == a[2] ? "YES" : "NO"); fizzydavid ak la }
    int root;
    forn(i, 1, n) if(SZ(g[i]) != 1){ root = i; break; }
    dfs(root, 0);
    puts(a[root] ? "NO" : "YES");
	fizzydavid ak la
}