/*
解説AC

g[i]の求め方.総積ではなく総和であることに注意

*/
#include<iostream>
#include<string>
#include<queue>
#include<vector>
#include<cassert>
#include<random>
#include<set>
#include<map>
#include<cassert>
#include<unordered_map>
#include<bitset>
#include<numeric>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf=1<<30;
const ll INF=1LL<<62;
typedef pair<int,ll> P;
typedef pair<int,P> PP; 
const ll MOD=998244353;


const int mod = 998244353;
struct mint {
  long long x; // typedef long long long long;
  mint(long long x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}//後ろのconstはメンバxを変更しないことを示す
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  bool operator==(const mint a) const {return a.x==x;}
  mint pow(unsigned long long int t) const {
    assert(t>=0);
    if (!t) return 1;
    //aがtで割り切れない場合には t%=(mod-1)をして良い
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};

std::istream& operator>>(std::istream& is, mint& a) { return is >> a.x;}
std::ostream& operator<<(std::ostream& os, const mint& a) { return os << a.x;}



struct combination {
  std::vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};



int main(){
    int N;
    cin>>N;
    vector<mint> A(N);
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    vector<vector<int>> G(N);
    for(int i=0;i<N-1;i++){
        int u,v;
        cin>>u>>v;
        u--;v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    vector<mint> g(N,0);
    auto dfs=[&](auto f,int now,int pre)->void{
        mint multi=0;
        bool is_leaf=true;
        for(int e:G[now]){
            if(e==pre) continue;
            f(f,e,now);
            is_leaf=false;
            multi+=g[e];
        }
        
        g[now]=(multi+1)*A[now];
        
        return;
    };
    dfs(dfs,0,-1);


    mint ans=0;

    auto dfs2=[&](auto f,int now,int pre)->void{
        mint sum_child=0;
        mint sum_square_child=0;
        bool is_leaf=true;
        for(int e:G[now]){
            if(e==pre) continue;
            f(f,e,now);
            is_leaf=false;
            sum_child+=g[e];
            sum_square_child+=g[e]*g[e];
        }
        
        ans+=A[now]*(sum_child + (sum_child*sum_child - sum_square_child)/2);
    };

    dfs2(dfs2,0,-1);
    cout<<ans<<endl;

}