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;
template <typename T>
void out(T res) {
  cout << res << endl;
  exit(0);
}
string a[55];
int n, m, k, st, fn;
bool allowed[26];
int way[55][55];
int dist[55][55];
int dr[] = {+1, 0, -1, 0};
int dc[] = {0, +1, 0, -1};
bool inBounds(int r, int c) { return (0 <= r && r < n && 0 <= c && c < m); }
bool Check(int r, int c) {
  return inBounds(r, c) &&
         (a[r][c] == 'S' || a[r][c] == 'T' || allowed[a[r][c] - 'a']);
}
string getWay(char c1, char c2, char c3, char c4) {
  memset(dist, -1, sizeof(dist));
  dist[fn / m][fn % m] = 0;
  queue<int> q;
  q.push(fn);
  while (!q.empty()) {
    int r = q.front() / m;
    int c = q.front() % m;
    q.pop();
    int nr, nc;
    for (int p = (0), _b(4); p < _b; p++)
      if (Check(nr = r + dr[p], nc = c + dc[p])) {
        if (dist[nr][nc] != -1) continue;
        dist[nr][nc] = dist[r][c] + 1;
        q.push(nr * m + nc);
      }
  }
  if (dist[st / m][st % m] == -1) return "-1";
  vector<int> v;
  v.push_back(st);
  string res = "";
  for (int i = (1), _b(dist[st / m][st % m]); i < _b; i++) {
    int nr, nc;
    int mm = 26;
    for (int j = (0), _b((int)(v).size()); j < _b; j++)
      for (int p = (0), _b(4); p < _b; p++)
        if (Check(nr = v[j] / m + dr[p], nc = v[j] % m + dc[p]))
          if (dist[nr][nc] == dist[st / m][st % m] - i)
            mm = min(mm, a[nr][nc] - 'a');
    res += (mm + 'a');
    vector<int> add;
    for (int j = (0), _b((int)(v).size()); j < _b; j++)
      for (int p = (0), _b(4); p < _b; p++)
        if (Check(nr = v[j] / m + dr[p], nc = v[j] % m + dc[p]))
          if (dist[nr][nc] == dist[st / m][st % m] - i)
            if (a[nr][nc] - 'a' == mm) add.push_back(nr * m + nc);
    v.clear();
    for (int j = (0), _b((int)(add).size()); j < _b; j++) v.push_back(add[j]);
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
  }
  return res;
}
int main() {
  cin >> n >> m >> k;
  for (int i = (0), _b(n); i < _b; i++) cin >> a[i];
  for (int i = (0), _b(n); i < _b; i++)
    for (int j = (0), _b(m); j < _b; j++) {
      if (a[i][j] == 'S') st = i * m + j;
      if (a[i][j] == 'T') fn = i * m + j;
    }
  string res = "-1";
  for (int c1 = (0), _b(26); c1 < _b; c1++)
    for (int c2 = (c1), _b(26); c2 < _b; c2++)
      for (int c3 = (c2), _b(26); c3 < _b; c3++)
        for (int c4 = (c3), _b(26); c4 < _b; c4++) {
          if (k > 0) allowed[c1] = true;
          if (k > 1) allowed[c2] = true;
          if (k > 2) allowed[c3] = true;
          if (k > 3) allowed[c4] = true;
          string curr = getWay(c1, c2, c3, c4);
          if (curr != "-1")
            if (res == "-1" || (int)(res).size() > (int)(curr).size() ||
                ((int)(res).size() == (int)(curr).size() && res > curr))
              res = curr;
          allowed[c1] = allowed[c2] = allowed[c3] = allowed[c4] = false;
        }
  out(res);
  return 0;
}