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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

#define INF 1001000100010001000
#define MOD 1000000007
#define EPS 1e-10
#define int long long
#define rep(i, N) for (int i = 0; i < N; i++)
#define Rep(i, N) for (int i = 1; i < N; i++)
#define For(i, a, b) for (int i = (a); i < (b); i++)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define pii pair<int, int>
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define vp vector< pii >
#define all(a) (a).begin(), (a).end()
#define Int(x) int x; cin >> x;
#define int2(x, y) Int(x); Int(y);
#define int3(x, y, z) Int(x); int2(y, z);
#define in(x, a, b) ((a) <= (x) && (x) < (b))
#define fir first
#define sec second
#define ffir first.first
#define fsec first.second
#define sfir second.first
#define ssec second.second
#define Decimal fixed << setprecision(10)

//int dxy[5] = {0, 1, 0, -1, 0};
// 01Trie BIT CHT DFT FFT avl cmd cusum dijkstra dinic geo2 gin graph kruskal lca lcm matrix ncm next_combination ppuf segtree st tmp topcoder uf vi 

using Weight = int;
using Flow = int;
struct Edge {
    int s, d; Weight w; Flow c;
    Edge() {};
    Edge(int s, int d, Weight w = 1) : s(s), d(d), w(w), c(w) {};
};
bool operator<(const Edge &e1, const Edge &e2) { return e1.w < e2.w; }
bool operator>(const Edge &e1, const Edge &e2) { return e2 < e1; }
inline ostream &operator<<(ostream &os, const Edge &e) { return (os << '(' << e.s << ", " << e.d << ", " << e.w << ')'); }

using Edges = vector<Edge>;
using Graph = vector<Edges>;
using Array = vector<Weight>;
using Matrix = vector<Array>;

void addArc(Graph &g, int s, int d, Weight w = 1) {
    g[s].emplace_back(s, d, w);
}
void addEdge(Graph &g, int a, int b, Weight w = 1) {
    addArc(g, a, b, w);
    addArc(g, b, a, w);
}

bool fail;

int dfs(Graph &g, vi &data, int pos, int bef)
{
    if (g[pos].size() == 1) {
        return data[pos];
    }
    int yui = 0;
    vi cnt;
    for (auto i : g[pos]) {
        if (i.d != bef) {
            cnt.pb(dfs(g, data, i.d, pos));
            yui += cnt.back();
        }
    }
    int kyoko = data[pos] * 2 - yui;
    if (max(*max_element(all(cnt)), yui/2) >= kyoko) {
        return kyoko;
    } else {
        fail = true;
        return -1;
    }
}

signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    Int(n);
    vector<int> data(n);
    for (int i = 0; i < n; i++) {
        cin >> data[i];
    }

    Graph g(n);
    for (int i = 0; i < n-1; i++) {
        int2(tmpx, tmpy);
        tmpx--; tmpy--;
        addEdge(g, tmpx, tmpy );
    }

    if (n == 2) {
        if (data[0] == data[1]) 
            std::cout << "YES" << std::endl;
        else 
            std::cout << "NO" << std::endl;
        return 0;
    }

    int p = 0;
    rep(i, n) {
        if (g[i].size() > 1) {
            p = i;
            break;
        }
    }

    fail = false;
    if (dfs(g, data, p, -1) == 0) {
        assert(false);
        std::cout << "YES" << std::endl;
    } else {
        std::cout << "NO" << std::endl;
    }

    return 0;
}