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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define _p(...) (void)printf(__VA_ARGS__)
#define forr(x,arr) for(auto&& x:arr)
#define _overload3(_1,_2,_3,name,...) name
#define _rep2(i,n) _rep3(i,0,n)
#define _rep3(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,_rep3,_rep2,)(__VA_ARGS__)
#define _rrep2(i,n) _rrep3(i,0,n)
#define _rrep3(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2,)(__VA_ARGS__)
#define all(x) (x).begin(), (x).end()
#define bit(n) (1LL<<(n))
#define sz(x) ((int)(x).size())
#define fst first
#define snd second
using ll=long long;
using pii=pair<int,int>;using pll=pair<ll,ll>;using pil=pair<int,ll>;using pli=pair<ll,int>;
using vs=vector<string>;using vvs=vector<vs>;using vvvs=vector<vvs>;
using vb=vector<bool>;using vvb=vector<vb>;using vvvb=vector<vvb>;
using vi=vector<int>;using vvi=vector<vi>;using vvvi=vector<vvi>;
using vl=vector<ll>;using vvl=vector<vl>;using vvvl=vector<vvl>;
using vd=vector<double>;using vvd=vector<vd>;using vvvd=vector<vvd>;
using vpii=vector<pii>;using vvpii=vector<vpii>;using vvvpii=vector<vvpii>;
template<class T,class U>istream&operator>>(istream&is,pair<T,U>&P){is>>P.fst;is>>P.snd;return is;}
template<class T>T read(){T t;cin>>t;return t;}
template<class T,class U>ostream&operator<<(ostream&o,const pair<T,U>&p){return o<<'('<<p.fst<<", "<<p.snd<<')';}
template<class T>ostream&operator<<(ostream&o,const vector<T>&t){o<<"{";rep(i,sz(t))o<<t[i]<<",";o<<"}"<<endl;return o;}
vs s_p_l_i_t(const string&s,char c){vs v;stringstream ss(s);string x;while(getline(ss,x,c))v.emplace_back(x);return move(v);}
void e_r_r(vs::iterator it){cerr<<endl;}
template<class T,class... Args>void e_r_r(vs::iterator it,T a,Args... args){cerr<<it->substr((*it)[0]==' ',it->length())<<" = "<<a<<", ";e_r_r(++it,args...);}
#define out(args...){vs a_r_g_s=s_p_l_i_t(#args,',');e_r_r(a_r_g_s.begin(),args);}

bool solve() {
  int n = read<int>();
  vector<int> A(n);
  rep(i, n) A[i] = read<int>();
  vvi E(n);
  rep(i, n-1) {
    int u = read<int>() - 1;
    int v = read<int>() - 1;
    E[u].push_back(v);
    E[v].push_back(u);
  }

  bool ans = true;
  vi C(n);
  {
    function<void(int, int)> dfs = [&](int cur, int pre) {
      int sum_c = 0;
      int c_max = 0;

      forr(nex, E[cur]) {
        if (nex != pre) {
          dfs(nex, cur);
          sum_c += C[nex];
          c_max = max(c_max, C[nex]);
        }
      }

      if (sz(E[cur]) == 1) {
        C[cur] = A[cur];
      }
      else {
        C[cur] = 2 * A[cur] - sum_c;
        if ((sum_c - C[cur]) % 2) ans = false;
        int l = (sum_c - C[cur]) / 2;
        if (l < 0) ans = false;
        if (c_max - C[cur] > l) ans = false;
      }
    };
    dfs(0, -1);
  }

  return ans;
}

void Main() {
  cout << (solve() ? "YES" : "NO") << endl;
}
int main() { cin.tie(nullptr); ios::sync_with_stdio(false); Main(); return 0; }