Problem: You already know that Valery's favorite sport is biathlon. Due to your help, he learned to shoot without missing, and his skills are unmatched at the shooting range. But now a smaller task is to be performed, he should learn to complete the path fastest.

The track's map is represented by a rectangle n × m in size divided into squares. Each square is marked with a lowercase Latin letter (which means the type of the plot), with the exception of the starting square (it is marked with a capital Latin letters S) and the terminating square (it is marked with a capital Latin letter T). The time of movement from one square to another is equal to 1 minute. The time of movement within the cell can be neglected. We can move from the cell only to side-adjacent ones, but it is forbidden to go beyond the map edges. Also the following restriction is imposed on the path: it is not allowed to visit more than k different types of squares (squares of one type can be visited an infinite number of times). Squares marked with S and T have no type, so they are not counted. But S must be visited exactly once — at the very beginning, and T must be visited exactly once — at the very end.

Your task is to find the path from the square S to the square T that takes minimum time. Among all shortest paths you should choose the lexicographically minimal one. When comparing paths you should lexicographically represent them as a sequence of characters, that is, of plot types.

Input

The first input line contains three integers n, m and k (1 ≤ n, m ≤ 50, n·m ≥ 2, 1 ≤ k ≤ 4). Then n lines contain the map. Each line has the length of exactly m characters and consists of lowercase Latin letters and characters S and T. It is guaranteed that the map contains exactly one character S and exactly one character T.

Pretest 12 is one of the maximal tests for this problem.

Output

If there is a path that satisfies the condition, print it as a sequence of letters — the plot types. Otherwise, print "-1" (without quotes). You shouldn't print the character S in the beginning and T in the end.

Note that this sequence may be empty. This case is present in pretests. You can just print nothing or print one "End of line"-character. Both will be accepted.

Examples

Input

5 3 2
Sba
ccc
aac
ccc
abT


Output

bcccc


Input

3 4 1
Sxyy
yxxx
yyyT


Output

xxxx


Input

1 3 3
TyS


Output

y


Input

1 4 1
SxyT


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int dy[4] = {0, 0, -1, 1};
const int dx[4] = {1, -1, 0, 0};
int n, m, k;
vector<string> s;
pair<int, string> answ;
bool us[26];
void get_way() {
  int d[50][50];
  bool used[50][50];
  pair<int, int> from[50][50];
  int sy, sx, ty, tx;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      d[i][j] = 1000000000;
      if (s[i][j] == 'S') {
        sy = i;
        sx = j;
      }
      if (s[i][j] == 'T') {
        ty = i;
        tx = j;
      }
      used[i][j] = false;
    }
  d[ty][tx] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(ty, tx));
  while (!q.empty()) {
    int cury = q.front().first;
    int curx = q.front().second;
    q.pop();
    for (int i = 0; i < 4; i++) {
      int toy = cury + dy[i];
      int tox = curx + dx[i];
      if (toy >= 0 && toy < n && tox >= 0 && tox < m && d[toy][tox] == INF) {
        if (s[toy][tox] != 'S' && s[toy][tox] != 'T' && !us[s[toy][tox] - 'a'])
          continue;
        d[toy][tox] = d[cury][curx] + 1;
        q.push(make_pair(toy, tox));
      }
    }
  }
  if (d[sy][sx] == INF) return;
  vector<pair<int, int> > st, st2;
  st.push_back(make_pair(sy, sx));
  used[sy][sx] = true;
  while (true) {
    for (int i = 0; i < st.size(); i++) {
      int cury = st[i].first;
      int curx = st[i].second;
      for (int j = 0; j < 4; j++) {
        int toy = cury + dy[j];
        int tox = curx + dx[j];
        if (toy >= 0 && toy < n && tox >= 0 && tox < m &&
            d[cury][curx] - d[toy][tox] == 1 && !used[toy][tox]) {
          used[toy][tox] = true;
          from[toy][tox] = make_pair(cury, curx);
          if (st2.size() == 0) {
            st2.push_back(make_pair(toy, tox));
          } else {
            if (s[toy][tox] < s[st2[0].first][st2[0].second]) {
              st2.clear();
              st2.push_back(make_pair(toy, tox));
            } else if (s[toy][tox] == s[st2[0].first][st2[0].second]) {
              st2.push_back(make_pair(toy, tox));
            }
          }
        }
      }
    }
    st = st2;
    st2.clear();
    if (st.size() == 0) break;
  }
  string ans = "";
  int cy = ty, cx = tx;
  while (true) {
    int tty = from[cy][cx].first;
    int ttx = from[cy][cx].second;
    if (tty == sy && ttx == sx) break;
    ans += s[tty][ttx];
    cy = tty;
    cx = ttx;
  }
  reverse(ans.begin(), ans.end());
  if (d[sy][sx] < answ.first) {
    answ = make_pair(d[sy][sx], ans);
  } else if (d[sy][sx] == answ.first && ans < answ.second) {
    answ = make_pair(d[sy][sx], ans);
  }
}
void rec(int cur, int cnt) {
  if (cnt == k) {
    get_way();
    return;
  }
  for (int i = cur; i < 26; i++) {
    us[i] = true;
    rec(i + 1, cnt + 1);
    us[i] = false;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    string str;
    cin >> str;
    s.push_back(str);
  }
  answ = make_pair(INF, "");
  rec(0, 0);
  if (answ.first == INF) {
    cout << -1;
  } else
    cout << answ.second;
  return 0;
}