long long calc(int ind, int used, int rem, bool change, int str, int used_l, const string & s, const string & p) {
  if (rem < 0)
    return 0;
  if (ind == p.size())
    return rem == 0 && !change;

  long long & r = dp[used][rem][change][str];
  if (r != -1)
    return r;
  r = 0;
  int letters = 0;
  char cur_l;

  for (int i = 0; i < (int)p.size(); ++i) {
    if (!((used >> i) & 1)) {
      cur_l = p[i];
      if ((letters >> (cur_l - 'a')) & 1) // don't reuse a char we used her before or u wilong long consider duplicates (cc gg case)
        continue;

      letters |= (1 << (cur_l - 'a'));
      r += calc(ind + 1, used | (1 << i), rem - abs(s[ind] - cur_l), change, str, used_l | (1 << (cur_l - 'a')), s, p);

      if (change) { // try 1 change for any
        if (!((used_l >> (cur_l - 1 - 'a')) & 1))  // don't reuse a char we used her before or u wilong long consider duplicates (cc gg case)
          r += calc(ind + 1, used | (1 << i), rem - abs(s[ind] - (cur_l - 1)), 0, str, used_l | (1 << (cur_l - 1 - 'a')), s, p);

        if (!((used_l >> (cur_l + 1 - 'a')) & 1))
          r += calc(ind + 1, used | (1 << i), rem - abs(s[ind] - (cur_l + 1)), 0, str, used_l | (1 << (cur_l + 1 - 'a')), s, p);
      }
    }
  }
  return r;
}