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
#include <bits/stdc++.h>
using namespace std;
const int M = 100 + 10;
char board[M][M];
char graph[M][M];
struct NODE {
  int parent_row;
  int parent_col;
  int dis;
  int visited;
  string par;
} node[M][M];
int cnt = 0;
int pattern[26 * 26 * 26 * 26][5];
void nchoosek(int num, int *a, int tar, int len, int isize, int spe) {
  if (isize == tar) {
    int j = 0;
    for (int i = 0; i < len; i++) {
      if (a[i] == 1) {
        pattern[cnt][j] = i;
        j++;
      }
    }
    cnt++;
    return;
  } else {
    for (int i = num; i < len; i++) {
      if (i != spe) {
        a[i] = 1;
        nchoosek(i + 1, a, tar, len, isize + 1, spe);
        a[i] = 0;
      }
    }
    return;
  }
}
string ans;
int LEN = 1E+9;
int dir[8] = {0, 1, 0, -1, -1, 0, 1, 0};
int n, m;
void BFS(int sr, int sc, int er, int ec) {
  queue<pair<int, int> > Q;
  Q.push(make_pair(sr, sc));
  node[sr][sc].visited = 1;
  node[sr][sc].dis = 0;
  while (!Q.empty()) {
    int row = Q.front().first;
    int col = Q.front().second;
    Q.pop();
    for (int i = 0; i < 8; i = i + 2) {
      int nrow = row + dir[i];
      int ncol = col + dir[i + 1];
      if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m &&
          (graph[nrow][ncol] == 'O' || graph[nrow][ncol] == 'T')) {
        if (node[nrow][ncol].visited == 1) {
          if (node[nrow][ncol].dis == node[row][col].dis + 1) {
            int prow = node[nrow][ncol].parent_row;
            int pcol = node[nrow][ncol].parent_col;
            if (node[row][col].par < node[prow][pcol].par) {
              node[nrow][ncol].parent_row = row;
              node[nrow][ncol].parent_col = col;
              node[nrow][ncol].par = node[row][col].par + board[nrow][ncol];
            }
          }
        } else {
          node[nrow][ncol].parent_row = row;
          node[nrow][ncol].parent_col = col;
          node[nrow][ncol].dis = node[row][col].dis + 1;
          node[nrow][ncol].visited = 1;
          node[nrow][ncol].par = node[row][col].par + board[nrow][ncol];
          if (nrow == er && ncol == ec) {
            ;
          } else {
            Q.push(make_pair(nrow, ncol));
          }
        }
      }
    }
  }
  if (node[er][ec].visited == 1) {
    int row = er;
    int col = ec;
    vector<char> tmp;
    while (row != sr || col != sc) {
      tmp.push_back(board[row][col]);
      int irow = node[row][col].parent_row;
      int icol = node[row][col].parent_col;
      row = irow;
      col = icol;
    }
    tmp.push_back(board[sr][sc]);
    if (tmp.size() < LEN) {
      ans = "";
      for (int i = tmp.size() - 1; i >= 0; i--) {
        ans = ans + tmp[i];
      }
      LEN = tmp.size();
    } else {
      if (tmp.size() == LEN) {
        string ss = "";
        for (int i = tmp.size() - 1; i >= 0; i--) {
          ss += tmp[i];
        }
        if (ss < ans) {
          ans = ss;
        }
      }
    }
  }
  return;
}
int main() {
  int k;
  cin >> n >> m >> k;
  set<char> letter;
  int sr, sc;
  int er, ec;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < m; j++) {
      board[i][j] = s[j];
      if (s[j] == 'S') {
        sr = i;
        sc = j;
      }
      if (s[j] == 'T') {
        er = i;
        ec = j;
      }
      if (s[j] != 'S' && s[j] != 'T') {
        letter.insert(s[j]);
      }
    }
  }
  vector<char> grid;
  for (set<char>::iterator it = letter.begin(); it != letter.end(); it++) {
    grid.push_back(*it);
  }
  sort(grid.begin(), grid.end());
  if (grid.size() < k) {
    k = grid.size();
  }
  for (int kk = 0; kk < 8; kk += 2) {
    int isr = sr + dir[kk];
    int isc = sc + dir[kk + 1];
    if (isr >= 0 && isr < n && isc >= 0 && isc < m) {
      for (int ksize = 1; ksize <= k; ksize++) {
        cnt = 0;
        int a[26];
        for (int j = 0; j < grid.size(); j++) {
          a[j] = 0;
        }
        char cc = board[isr][isc];
        int inum = -1;
        for (int ii = 0; ii < grid.size(); ii++) {
          if (grid[ii] == cc) {
            inum = ii;
            break;
          }
        }
        a[inum] = 1;
        nchoosek(0, a, ksize, grid.size(), 1, inum);
        for (int i = 0; i < cnt; i = i + 1) {
          for (int row = 0; row < n; row++) {
            for (int col = 0; col < m; col++) {
              if (board[row][col] == 'S') {
                graph[row][col] = 'S';
              }
              if (board[row][col] == 'T') {
                graph[row][col] = 'T';
              }
              if (board[row][col] != 'S' && board[row][col] != 'T') {
                graph[row][col] = 'X';
                for (int j = 0; j < ksize; j++) {
                  if (board[row][col] == grid[pattern[i][j]]) {
                    graph[row][col] = 'O';
                  }
                }
              }
            }
          }
          for (int row = 0; row < n; row++) {
            for (int col = 0; col < m; col++) {
              node[row][col].visited = 0;
            }
          }
          for (int row = 0; row < n; row++) {
            for (int col = 0; col < m; col++) {
              node[row][col].visited = 0;
            }
          }
          BFS(isr, isc, er, ec);
        }
      }
    }
  }
  if (LEN == 1E+9) {
    if (n * m > 2) cout << "-1" << endl;
  } else {
    for (int i = 0; i < ans.size() - 1; i++) {
      cout << ans[i];
    }
    cout << endl;
  }
  return 0;
}
