description: For a sequence of strings [t_1, t_2, ..., t_m], let's define the function f([t_1, t_2, ..., t_m]) as the number of different strings (including the empty string) that are subsequences of at least one string t_i. f([]) = 0 (i. e. the number of such strings for an empty sequence is 0).
  
  You are given a sequence of strings [s_1, s_2, ..., s_n]. Every string in this sequence consists of lowercase Latin letters and is sorted (i. e., each string begins with several (maybe zero) characters a, then several (maybe zero) characters b, ..., ends with several (maybe zero) characters z).
  
  For each of 2^n subsequences of [s_1, s_2, ..., s_n], calculate the value of the function f modulo 998244353.
  
  Input
  
  The first line contains one integer n (1 ≤ n ≤ 23) — the number of strings.
  
  Then n lines follow. The i-th line contains the string s_i (1 ≤ |s_i| ≤ 2 ⋅ 10^4), consisting of lowercase Latin letters. Each string s_i is sorted.
  
  Output
  
  Since printing up to 2^{23} integers would be really slow, you should do the following:
  
  For each of the 2^n subsequences (which we denote as [s_{i_1}, s_{i_2}, ..., s_{i_k}]), calculate f([s_{i_1}, s_{i_2}, ..., s_{i_k}]), take it modulo 998244353, then multiply it by k ⋅ (i_1 + i_2 + ... + i_k). Print the XOR of all 2^n integers you get.
  
  The indices i_1, i_2, ..., i_k in the description of each subsequences are 1-indexed (i. e. are from 1 to n).
  
  Examples
  
  Input
  
  
  3
  a
  b
  c
  
  
  Output
  
  
  92
  
  
  Input
  
  
  2
  aa
  a
  
  
  Output
  
  
  21
  
  
  Input
  
  
  2
  a
  a
  
  
  Output
  
  
  10
  
  
  Input
  
  
  2
  abcd
  aabb
  
  
  Output
  
  
  124
  
  
  Input
  
  
  3
  ddd
  aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
  aaaaaaaabbbbbbbbbbbcccccccccccciiiiiiiiiiiiiiiiiiiiiiooooooooooqqqqqqqqqqqqqqqqqqvvvvvzzzzzzzzzzzz
  
  
  Output
  
  
  15706243380
solution: #include <bits/stdc++.h>
  const double PI = 3.1415926535897932384626433;
  using namespace std;
  struct edge {
    long long to, cost;
    edge() {}
    edge(long long a, long long b) { to = a, cost = b; }
  };
  const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
  const int mod = 998244353;
  struct mint {
    int x;
    mint(long long y = 0) {
      if (y < 0 || y >= mod) y = (y % mod + mod) % mod;
      x = y;
    }
    mint(const mint &ope) { x = ope.x; }
    mint operator-() { return mint(-x); }
    mint operator+(const mint &ope) { return mint(x) += ope; }
    mint operator-(const mint &ope) { return mint(x) -= ope; }
    mint operator*(const mint &ope) { return mint(x) *= ope; }
    mint operator/(const mint &ope) { return mint(x) /= ope; }
    mint &operator+=(const mint &ope) {
      x += ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint &operator-=(const mint &ope) {
      x += mod - ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint &operator*=(const mint &ope) {
      long long tmp = x;
      tmp *= ope.x, tmp %= mod;
      x = tmp;
      return *this;
    }
    mint &operator/=(const mint &ope) {
      long long n = mod - 2;
      mint mul = ope;
      while (n) {
        if (n & 1) *this *= mul;
        mul *= mul;
        n >>= 1;
      }
      return *this;
    }
    mint inverse() { return mint(1) / *this; }
    bool operator==(const mint &ope) { return x == ope.x; }
    bool operator!=(const mint &ope) { return x != ope.x; }
    bool operator<(const mint &ope) const { return x < ope.x; }
  };
  mint modpow(mint a, long long n) {
    if (n == 0) return mint(1);
    if (n % 2)
      return a * modpow(a, n - 1);
    else
      return modpow(a * a, n / 2);
  }
  istream &operator>>(istream &is, mint &ope) {
    long long t;
    is >> t, ope.x = t;
    return is;
  }
  ostream &operator<<(ostream &os, mint &ope) { return os << ope.x; }
  ostream &operator<<(ostream &os, const mint &ope) { return os << ope.x; }
  long long modpow(long long a, long long n, long long mod) {
    if (n == 0) return 1;
    if (n % 2)
      return ((a % mod) * (modpow(a, n - 1, mod) % mod)) % mod;
    else
      return modpow((a * a) % mod, n / 2, mod) % mod;
  }
  vector<mint> fact, fact_inv;
  void make_fact(int n) {
    fact.resize(n + 1), fact_inv.resize(n + 1);
    fact[0] = mint(1);
    for (long long i = (1); (i) <= (n); (i)++) fact[i] = fact[i - 1] * mint(i);
    fact_inv[n] = fact[n].inverse();
    for (long long i = (n - 1); (i) >= (0); (i)--)
      fact_inv[i] = fact_inv[i + 1] * mint(i + 1);
  }
  mint comb(int n, int k) {
    if (n < 0 || k < 0 || n < k) return mint(0);
    return fact[n] * fact_inv[k] * fact_inv[n - k];
  }
  mint perm(int n, int k) { return comb(n, k) * fact[k]; }
  vector<int> prime, pvec, qrime;
  void make_prime(int n) {
    prime.resize(n + 1);
    for (long long i = (2); (i) <= (n); (i)++) {
      if (prime[i] == 0) pvec.push_back(i), prime[i] = i;
      for (auto p : pvec) {
        if (i * p > n || p > prime[i]) break;
        prime[i * p] = p;
      }
    }
  }
  void make_qrime(int n) {
    qrime.resize(n + 1);
    for (long long i = (2); (i) <= (n); (i)++) {
      int ni = i / prime[i];
      if (prime[i] == prime[ni])
        qrime[i] = qrime[ni] * prime[i];
      else
        qrime[i] = prime[i];
    }
  }
  bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
  void mark() { cout << "*" << endl; }
  void yes() { cout << "YES" << endl; }
  void no() { cout << "NO" << endl; }
  long long floor(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return a / b;
    else
      return -((-a + b - 1) / b);
  }
  long long ceil(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return (a + b - 1) / b;
    else
      return -((-a) / b);
  }
  long long modulo(long long a, long long b) {
    b = abs(b);
    return a - floor(a, b) * b;
  }
  long long sgn(long long x) {
    if (x > 0) return 1;
    if (x < 0) return -1;
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long arith(long long x) { return x * (x + 1) / 2; }
  long long arith(long long l, long long r) { return arith(r) - arith(l - 1); }
  long long digitnum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret++;
    return ret;
  }
  long long digitsum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret += x % b;
    return ret;
  }
  string lltos(long long x) {
    string ret;
    for (; x; x /= 10) ret += x % 10 + '0';
    reverse((ret).begin(), (ret).end());
    return ret;
  }
  long long stoll(string &s) {
    long long ret = 0;
    for (auto c : s) ret *= 10, ret += c - '0';
    return ret;
  }
  template <typename T>
  void uniq(T &vec) {
    sort((vec).begin(), (vec).end());
    vec.erase(unique((vec).begin(), (vec).end()), vec.end());
  }
  int popcount(unsigned long long x) {
    x -= ((x >> 1) & 0x5555555555555555ULL),
        x = (x & 0x3333333333333333ULL) + ((x >> 2) & 0x3333333333333333ULL);
    return (((x + (x >> 4)) & 0x0F0F0F0F0F0F0F0FULL) * 0x0101010101010101ULL) >>
           56;
  }
  template <class S, class T>
  pair<S, T> &operator+=(pair<S, T> &s, const pair<S, T> &t) {
    s.first += t.first, s.second += t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T> &operator-=(pair<S, T> &s, const pair<S, T> &t) {
    s.first -= t.first, s.second -= t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first + t.first, s.second + t.second);
  }
  template <class S, class T>
  pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) {
    return pair<S, T>(s.first - t.first, s.second - t.second);
  }
  template <class T>
  T dot(const pair<T, T> &s, const pair<T, T> &t) {
    return s.first * t.first + s.second * t.second;
  }
  template <class T>
  T cross(const pair<T, T> &s, const pair<T, T> &t) {
    return s.first * t.second - s.second * t.first;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> &vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, deque<T> &deq) {
    for (long long i = 0; (i) < (long long)(deq).size(); (i)++)
      os << deq[i] << " ";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, pair<T, U> &ope) {
    os << "(" << ope.first << ", " << ope.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, const pair<T, U> &ope) {
    os << "(" << ope.first << ", " << ope.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, map<T, U> &ope) {
    for (auto p : ope) os << "(" << p.first << ", " << p.second << "),";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> &ope) {
    for (auto x : ope) os << x << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> &ope) {
    for (auto x : ope) os << x << " ";
    return os;
  }
  template <typename T>
  void outa(T a[], long long s, long long t) {
    for (long long i = (s); (i) <= (t); (i)++) {
      cout << a[i];
      if (i < t) cout << " ";
    }
    cout << endl;
  }
  void dump_func() { cout << endl; }
  template <class Head, class... Tail>
  void dump_func(Head &&head, Tail &&...tail) {
    cout << head;
    if (sizeof...(Tail) > 0) cout << " ";
    dump_func(std::move(tail)...);
  }
  long long n;
  long long cnt[23][26];
  mint sum[1 << 23];
  template <typename T>
  void zeta_transform(T a[], int n) {
    for (long long i = (0); (i) <= (n - 1); (i)++)
      for (long long j = (0); (j) <= ((1 << n) - 1); (j)++)
        if (j & (1 << i)) a[j] += a[j ^ (1 << i)];
  }
  int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    string s;
    for (long long i = (0); (i) <= (n - 1); (i)++) {
      cin >> s;
      for (auto c : s) cnt[i][c - 'a']++;
    }
    long long N = 1 << n;
    for (long long i = (1); (i) <= (N - 1); (i)++) {
      long long mn[26];
      for (long long j = (0); (j) <= (25); (j)++) mn[j] = 2e18;
      for (long long j = (0); (j) <= (n - 1); (j)++) {
        if ((i & (1 << j)) == 0) continue;
        for (long long k = (0); (k) <= (25); (k)++)
          (mn[k]) = min((mn[k]), (cnt[j][k]));
      }
      sum[i] = 1;
      for (long long j = (0); (j) <= (25); (j)++) sum[i] *= (mn[j] + 1);
      if (popcount(i) % 2 == 0) sum[i] *= -1;
    }
    zeta_transform(sum, n);
    long long ans = 0;
    for (long long i = (1); (i) <= (N - 1); (i)++) {
      long long tmp = 0;
      for (long long j = (0); (j) <= (n - 1); (j)++) {
        if (i & (1 << j)) tmp += j + 1;
      }
      tmp *= popcount(i);
      tmp *= sum[i].x;
      ans ^= tmp;
    }
    dump_func(ans);
    return 0;
  }
  