Problem: Facetook is a well known social network website, and it will launch a new feature called Facetook Priority Wall. This feature will sort all posts from your friends according to the priority factor (it will be described).

This priority factor will be affected by three types of actions: 

  * 1. "X posted on Y's wall" (15 points), 
  * 2. "X commented on Y's post" (10 points), 
  * 3. "X likes Y's post" (5 points). 



X and Y will be two distinct names. And each action will increase the priority factor between X and Y (and vice versa) by the above value of points (the priority factor between X and Y is the same as the priority factor between Y and X).

You will be given n actions with the above format (without the action number and the number of points), and you have to print all the distinct names in these actions sorted according to the priority factor with you.

Input

The first line contains your name. The second line contains an integer n, which is the number of actions (1 ≤ n ≤ 100). Then n lines follow, it is guaranteed that each one contains exactly 1 action in the format given above. There is exactly one space between each two words in a line, and there are no extra spaces. All the letters are lowercase. All names in the input will consist of at least 1 letter and at most 10 small Latin letters.

Output

Print m lines, where m is the number of distinct names in the input (excluding yourself). Each line should contain just 1 name. The names should be sorted according to the priority factor with you in the descending order (the highest priority factor should come first). If two or more names have the same priority factor, print them in the alphabetical (lexicographical) order.

Note, that you should output all the names that are present in the input data (excluding yourself), even if that person has a zero priority factor.

The lexicographical comparison is performed by the standard "<" operator in modern programming languages. The line a is lexicographically smaller than the line b, if either a is the prefix of b, or if exists such an i (1 ≤ i ≤ min(|a|, |b|)), that ai < bi, and for any j (1 ≤ j < i) aj = bj, where |a| and |b| stand for the lengths of strings a and b correspondently.

Examples

Input

ahmed
3
ahmed posted on fatma's wall
fatma commented on ahmed's post
mona likes ahmed's post


Output

fatma
mona


Input

aba
1
likes likes posted's post


Output

likes
posted

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("O2")
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr long long INF = 1999999999999999997;
constexpr int inf = INT_MAX;
constexpr int MAXSIZE = int(1e6) + 5;
constexpr auto PI = 3.14159265358979323846L;
constexpr auto oo = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(char c) { return string({c}); }
template <size_t N>
string to_string(bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgs() { ; }
template <typename Heads, typename... Tails>
void dbgs(Heads H, Tails... T) {
  cout << to_string(H) << " | ";
  dbgs(T...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string name;
  cin >> name;
  long long n;
  cin >> n;
  map<string, long long> points;
  string a, b, c, d, e;
  while (n--) {
    cin >> a >> b >> c >> d;
    if (c == "on") {
      cin >> e;
      if (a == name) {
        if (b == "posted")
          points[d.substr(0, d.size() - 2)] += 15;
        else
          points[d.substr(0, d.size() - 2)] += 10;
      } else if ((d.substr(0, d.size() - 2)) == name) {
        if (b == "posted")
          points[a] += 15;
        else
          points[a] += 10;
      } else {
        points[d.substr(0, d.size() - 2)] += 0;
        points[a] += 0;
      }
    } else {
      if (a == name) {
        points[c.substr(0, c.size() - 2)] += 5;
      } else if ((c.substr(0, c.size() - 2)) == name)
        points[a] += 5;
      else {
        points[a] += 0;
        points[c.substr(0, c.size() - 2)] += 0;
      }
    }
  }
  vector<string> ans2;
  vector<pair<long long, string>> ans;
  for (auto i : points) {
    if (i.second == 0)
      ans2.push_back(i.first);
    else
      ans.push_back(make_pair(-i.second, i.first));
  }
  sort(ans.begin(), ans.end());
  for (auto i : ans) cout << i.second << "\n";
  for (auto i : ans2) cout << i << "\n";
}