#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto& i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll

using namespace std;
using namespace atcoder;

using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}}
#define dame { cout << "No" << endl; return;}
#define dame1 { cout << -1 << endl; return;}
#define cout2(x,y) cout << x << " " << y << endl;
#define coutp(p) cout << p.fi << " " << p.se << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define coutv2(v) fore(vy , v) cout << vy << "\n";
#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define coutvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = dynamic_modint<0>;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define yesr { cout <<"Yes"<<endl; return;}
#define no {cout <<"No"<<endl;}
#define nor { cout <<"No"<<endl; return;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#define show(x) cerr<<#x<<" = "<<x<<endl
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show2(x,y) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<endl
#define show3(x,y,z) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<endl
#define show4(x,y,z,x2) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<" : "<<#x2<<" = "<<x2<<endl
#define test(x) cout << "test" << x << endl
#define showv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) fore(vy , v) cout << vy << "\n";
#define showvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) fore(vy , v) cout << vy.val() << "\n";
#else
#define show(x)
#define showp(p)
#define show2(x,y)
#define show3(x,y,z)
#define show4(x,y,z,x2)
#define test(x)
#define showv(v)
#define showv2(v)
#define showvm(v)
#define showvm2(v)
#endif

vs a;
mint ans;

template<typename T>
struct ReRooting {

    using F1 = function<T(T, T)>;
    using F2 = function<T()>;

    int n; // 頂点数
    vvi e,tmp_edge; // e:edge情報 親は最後に入る
    vector<int> d;
    vector<vector<T>> dp; // dp[x][i]、iはeに連動
    vector<vector<T>> lsum, rsum; // 左(右)からの累積和 
    vector<T> ans2;
    F1 op; // マージ演算関数
    F2 id; // 単位元

    ReRooting(int n,F1 op,F2 id) : n(n),e(n),tmp_edge(n),dp(n),lsum(n),rsum(n),ans2(n),op(op),id(id),d(n){}

    void add_edge(int u, int v) {
        tmp_edge[u].pb(v);
        tmp_edge[v].pb(u);
    }

    T dfs1(int x, int pre) {
        T ret = id();
        int i = 0;

        d[x] = 1;
    
        fore(y , tmp_edge[x]){
            if (y == pre) continue;
        
            dp[x].pb(dfs1(y , x));
            d[x] += d[y];
            e[x].pb(y);
            ret = op(ret , dp[x][i]);
        
            lsum[x].pb(id());
            rsum[x].pb(id());
        
            i++;
        }

        // ll l = stoll(a[x]);
        ret++;
        // ans += mint(l)*ret;
        ret *= mint(10).pow(sz(a[x]));
    
        if (pre != -1){
            e[x].pb(pre);
            dp[x].pb(id());
            lsum[x].pb(id());
            rsum[x].pb(id());
        }
    
        return ret+id();
    }

    void dfs2(int x,int pre,T preval) {
        int m = sz(e[x]);

        ll l = stoll(a[x]);
        if (pre!=-1) ans += preval*l*d[x];
    
        if (pre != -1) dp[x][m-1] = preval;
    
        rep(i,m){
            lsum[x][i] = dp[x][i];
            if (i != 0) lsum[x][i] = op(lsum[x][i], lsum[x][i-1]);
        }
    
        rrep(i,m){
            rsum[x][i] = dp[x][i];
            if (i != m-1) rsum[x][i] = op(rsum[x][i], rsum[x][i+1]);
        }
    
        if (m >= 1) ans2[x] = lsum[x][m-1];
        else ans2[x] = id();
    
        rep(i,m){
            int y = e[x][i];
            if (y == pre) continue;
            ans += dp[x][i]*l*(n-d[y]);
        
            T nxtval = id();
            if (i != 0) nxtval = op(nxtval , lsum[x][i-1]);
            if (i != m-1) nxtval = op(nxtval , rsum[x][i+1]);
            nxtval++;
            nxtval *= mint(10).pow(sz(a[x]));
        
            dfs2(y , x , nxtval+id());
        }
        ans += mint(l)*n;

    
        return;
    }

    void solve(){
        dfs1(0,-1);
        dfs2(0,-1,id());
    }
};

// 全方位木dp
// ReRooting<T> rr(n,op,id) : 初期化。nは頂点数、opはmerge演算関数、idは単位元
// op,idはラムダ式で定義
// add_edge(a,b) a,b間に辺を引く
// solve() dp実行、ansに答えが入る。

// dfs内は適宜変更すること

mint op(mint a,mint b){
    return a+b;
}
mint id(){
    return 0;
}


void solve(){
    int n; cin>>n;
    a.resize(n);
    ans = 0;
    rep(i,n) cin>>a[i];
    // auto op = [](mint a,mint b)->mint{
    //     return a+b;
    // };
    // auto id = []()->mint{
    //     return mint(0);
    // };
    ReRooting<mint> rr(n,op,id);
    rep(i,n-1){
        int a,b; cin>>a>>b;
        a--; b--;
        rr.add_edge(a,b);
    }
    rr.solve();
    outm;
 
    return;
}

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

    int t = 1;
    //cin>>t;

    rep(i,t){
        solve();
    }

    return 0;
}