Problem: While exploring the old caves, researchers found a book, or more precisely, a stash of mixed pages from a book. Luckily, all of the original pages are present and each page contains its number. Therefore, the researchers can reconstruct the book.

After taking a deeper look into the contents of these pages, linguists think that this may be some kind of dictionary. What's interesting is that this ancient civilization used an alphabet which is a subset of the English alphabet, however, the order of these letters in the alphabet is not like the one in the English language.

Given the contents of pages that researchers have found, your task is to reconstruct the alphabet of this ancient civilization using the provided pages from the dictionary.

Input

First-line contains two integers: n and k (1 ≤ n, k ≤ 10^3) — the number of pages that scientists have found and the number of words present at each page. Following n groups contain a line with a single integer p_i (0 ≤ n < 10^3) — the number of i-th page, as well as k lines, each line containing one of the strings (up to 100 characters) written on the page numbered p_i.

Output

Output a string representing the reconstructed alphabet of this ancient civilization. If the book found is not a dictionary, output "IMPOSSIBLE" without quotes. In case there are multiple solutions, output any of them.

Example

Input


3 3
2
b
b
bbac
0
a
aca
acba
1
ab
c
ccb


Output


acb

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
template <class T>
using PR = pair<T, T>;
const int MOD = 1e9 + 7;
const int mxN = 1e3 + 5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int tSZ = (1 << 21);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
constexpr int pcnt(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return 31 - __builtin_clz(x); }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
ll add(ll first, ll second) {
  first += second;
  if (first >= MOD) first %= MOD;
  if (first < 0) first += MOD;
  return first;
}
ll mul(ll first, ll second) {
  first *= second;
  if (first >= MOD) first %= MOD;
  return first;
}
ll pw(ll x, ll y) {
  ll ans = 1, ml = x;
  while (y) {
    if (y & 1) ans = mul(ans, ml);
    ml = mul(ml, ml);
    y >>= 1;
  }
  return ans;
}
ll inv(ll x) { return pw(x, MOD - 2); }
template <class T>
void remDup(vector<T>& x) {
  sort(begin(x), end(x));
  x.erase(unique(begin(x), end(x)), end(x));
}
template <class T, class U>
void erase(T& x, const U& v) {
  auto it = x.find(v);
  assert(it != end(x));
  x.erase(v);
}
template <class T>
void re(complex<T>& c);
template <class T, class U>
void re(pair<T, U>& p);
template <class T>
void re(vector<T>& v);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(db& d) {
  str t;
  re(t);
  d = stod(t);
}
void re(ld& d) {
  str t;
  re(t);
  d = stold(t);
}
template <class T, class... U>
void re(T& t, U&... u) {
  re(t);
  re(u...);
}
template <class T>
void re(complex<T>& c) {
  T a, b;
  re(a, b);
  c = {a, b};
}
template <class T, class U>
void re(pair<T, U>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& x) {
  for (auto& a : x) re(a);
}
template <class T, size_t SZ>
void re(array<T, SZ>& x) {
  for (auto& a : x) re(a);
}
template <class T>
void rv(int& n, vector<T>& x) {
  re(n);
  x.resize(n);
  for (auto& a : x) re(a);
}
str to_string(char c) { return str(1, c); }
str to_string(const char* second) { return (str)second; }
str to_string(str second) { return second; }
str to_string(bool b) { return to_string((int)b); }
template <class T>
str to_string(complex<T> c) {
  stringstream ss;
  ss << c;
  return ss.str();
}
str to_string(vector<bool> v) {
  str res = "{";
  for (int i = (0); i < ((int)(v).size()); i++) res += char('0' + v[i]);
  res += "}";
  return res;
}
template <size_t SZ>
str to_string(bitset<SZ> b) {
  str res = "";
  for (int i = (0); i < (SZ); i++) res += char('0' + b[i]);
  return res;
}
template <class T, class U>
str to_string(pair<T, U> p);
template <class T>
str to_string(T v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <class T, class U>
str to_string(pair<T, U> p) {
  return to_string(p.first) + " " + to_string(p.second);
}
template <class T>
void pr(T v) {
  cout << to_string(v);
}
template <class T, class... U>
void pr(const T& t, const U&... u) {
  pr(t);
  pr(u...);
}
void ps() { pr("\n"); }
template <class T, class... U>
void ps(const T& t, const U&... u) {
  pr(t);
  if (sizeof...(u)) pr(" ");
  ps(u...);
}
void setIn(str second) { freopen(second.c_str(), "r", stdin); }
void setOut(str second) { freopen(second.c_str(), "w", stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str st = "") {
  unsyncIO();
  if ((int)(st).size()) {
    setIn(st + ".in"), setOut(st + ".out");
  }
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
  cerr << to_string(t);
  if (sizeof...(u)) cerr << ", ";
  DBG(u...);
}
int n, k;
str matr[mxN * mxN];
vector<str> vv;
int graph[30][30];
int c_br[mxN * mxN];
bool ok = 1;
void calc() {
  for (int i = (0); i < ((int)(vv).size() - 1); i++) {
    str tmp = vv[i];
    str nxt = vv[i + 1];
    int k = 0;
    while (k < min((int)(tmp).size(), (int)(nxt).size()) && tmp[k] == nxt[k]) {
      k++;
    }
    if (k < min((int)(tmp).size(), (int)(nxt).size())) {
      graph[tmp[k] - 'a'][nxt[k] - 'a'] = 1;
    } else if ((int)(tmp).size() > (int)(nxt).size()) {
      ok = 0;
    }
  }
}
void solve() {
  vi vis(27, 0);
  set<int> alphabet;
  for (int i = (0); i < (n * k); i++) {
    if ((int)(matr[i]).size() != 0) {
      vv.push_back(matr[i]);
      for (auto& x : matr[i]) alphabet.insert(x - 'a');
    }
  }
  for (int i = (0); i < ((int)(vv).size()); i++) {
    c_br[i] = (int)(vv[i]).size();
  }
  calc();
  vi ans;
  for (int i = (0); i < (27); i++) {
    for (int j = (0); j < (27); j++) {
      if (vis[j] || !alphabet.count(j)) continue;
      int cnt = 0;
      for (int z = (0); z < (27); z++) {
        cnt += graph[z][j];
      }
      if (cnt == 0) {
        ans.push_back(j);
        vis[j] = 1;
        for (int z = (0); z < (27); z++) {
          graph[j][z] = 0;
        }
      }
    }
  }
  if (!ok || (int)(ans).size() != (int)(alphabet).size()) {
    ps("IMPOSSIBLE");
    return;
  }
  for (auto& x : ans) {
    pr(char(x + 'a'));
  }
  ps();
}
int main() {
  setIO();
  re(n, k);
  int c;
  for (int i = (0); i < (n); i++) {
    re(c);
    for (int j = (0); j < (k); j++) {
      re(matr[c * n + j]);
    }
  }
  solve();
  return 0;
}