/* -*- coding: utf-8 -*-
 *
 * 3363.cc:  No.3363 Two Closest Numbers - yukicoder
 */

#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

/* constant */

const int MAX_N = 200000;
const int MOD = 998244353;

/* typedef */

using vi = vector<int>;

/* global variables */

int cs[MAX_N], es[10];

/* subroutines */

void mknums(int es[], vi &a, vi &b) {
  int ds[10];
  copy(es, es + 10, ds);

  for (int i = 0, j = 9; i <= j;) {
    while (i < 10 && ds[i] == 0) i++;
    while (j > 0 && ds[j] == 0) j--;
    if (i > j) break;
    a.push_back(i), ds[i]--;
    b.push_back(j), ds[j]--;
  }
}

vi subv(vi &a, vi &b) { // a >= b
  int n = a.size();
  vi c(n);

  for (int i = n - 1, co = 0; i >= 0; i--) {
    c[i] = a[i] - b[i] - co;
    if (c[i] < 0) c[i] += 10, co = 1;
    else co = 0;
  }

  return c;
}

bool gtv(vi &a, vi &b) {
  int al = a.size(), bl = b.size(), l = max(al, bl);
  for (int i = al - l, j = bl - l; i < al; i++, j++) {
    int da = (i >= 0) ? a[i] : 0;
    int db = (j >= 0) ? b[j] : 0;
    if (da != db) return (da > db);
  }
  return false;
}

vi min_even(int es[]) {
  int m = 0;
  for (int i = 1; i <= 9; i++) m += es[i];
  if (m == 0) return vi();

  vi minc(m / 2, 9);
  for (int i = 1; i <= 9; i++)
    if (es[i] > 0)
      for (int j = i + 1; j <= 9; j++)
	if (es[j] > 0) {
	  es[i]--, es[j]--;
	  vi a(1, j), b(1, i);
	  mknums(es, a, b);
	  es[i]++, es[j]++;
	  
	  vi c = subv(a, b);
	  if (gtv(minc, c)) minc = c;
	  break;
	}
  return minc;
}

void printv(vi &a) {
  for (auto d: a) printf("%d", d);
  putchar('\n');
}

/* main */

int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", cs + i);

  for (int i = 0; i < n; i++) es[cs[i]]++;
  //for (int i = 1; i <= 9; i++) printf(" %d", es[i]); putchar('\n');

  vi minc;
  if (n & 1) {
    int mn = 1;
    while (es[mn] == 0) mn++;
    vi a, b;
    a.push_back(mn); es[mn]--;
    b.push_back(0);
    mknums(es, a, b);
    minc = subv(a, b);
    //printv(a); printv(b);
  }
  else {
    int ees[10];
    for (int i = 1; i <= 9; i++) ees[i] = (es[i] & 1);
    minc = min_even(ees);
    //printv(minc);

    for (int i = 1; i <= 9; i++)
      if (es[i] >= 2) {
	ees[i] += 2;
	auto c = min_even(ees);
	//printv(c);
	if (gtv(minc, c)) minc = c;
	ees[i] -= 2;
      }
  }
  //printv(minc);

  int sum = 0;
  for (auto d: minc) sum = (sum * 10LL + d) % MOD;

  printf("%d\n", sum);
  
  return 0;
}