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;
queue<pair<int, int> > q;
string s, jwb;
pair<int, int> u, aw, ak;
int r, c, bts, a, h, mi, ni, nj, ui, uj;
char sc[55][55];
int boleh[27];
string kjg[55][55];
int ari[4] = {-1, 0, 1, 0};
int arj[4] = {0, 1, 0, -1};
int main() {
  scanf("%d%d%d", &r, &c, &bts);
  for (int i = 0; i < r; i++) {
    scanf("%s", sc[i]);
    for (int j = 0; j < c; j++) {
      if (sc[i][j] == 'S') {
        aw = make_pair(i, j);
      }
      if (sc[i][j] == 'T') {
        ak = make_pair(i, j);
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    boleh[i] = 1;
  }
  for (int i = 0; i < bts; i++) {
    boleh[i] = 0;
  }
  jwb = "";
  mi = 2123123123;
  do {
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) {
        kjg[i][j] = "";
      }
    }
    q.push(aw);
    while (!q.empty()) {
      ui = q.front().first;
      uj = q.front().second;
      q.pop();
      for (int k = 0; k < 4; k++) {
        ni = ui + ari[k];
        nj = uj + arj[k];
        if ((0 <= ni) && (ni < r) && (0 <= nj) && (nj < c) &&
            (sc[ni][nj] != 'S')) {
          if (sc[ni][nj] == 'T') {
            if (kjg[ni][nj].size() == 0) {
              kjg[ni][nj] = kjg[ui][uj] + sc[ni][nj];
              q.push(make_pair(ni, nj));
            } else if ((kjg[ni][nj].size() == kjg[ui][uj].size() + 1) &&
                       (kjg[ni][nj] > kjg[ui][uj] + sc[ni][nj])) {
              kjg[ni][nj] = kjg[ui][uj] + sc[ni][nj];
            }
          } else if (boleh[sc[ni][nj] - 'a'] == 0) {
            if (kjg[ni][nj].size() == 0) {
              kjg[ni][nj] = kjg[ui][uj] + sc[ni][nj];
              q.push(make_pair(ni, nj));
            } else if ((kjg[ni][nj].size() == kjg[ui][uj].size() + 1) &&
                       (kjg[ni][nj] > kjg[ui][uj] + sc[ni][nj])) {
              kjg[ni][nj] = kjg[ui][uj] + sc[ni][nj];
            }
          }
        }
      }
    }
    if ((int)kjg[ak.first][ak.second].size() > 0) {
      kjg[ak.first][ak.second].erase(kjg[ak.first][ak.second].size() - 1, 1);
      if ((int)kjg[ak.first][ak.second].size() < mi) {
        mi = kjg[ak.first][ak.second].size();
        jwb = kjg[ak.first][ak.second];
      } else if ((int)kjg[ak.first][ak.second].size() == mi) {
        jwb = min(jwb, kjg[ak.first][ak.second]);
      }
    }
  } while (next_permutation(boleh, boleh + 26));
  if (mi == 2123123123) {
    cout << -1 << endl;
  } else {
    cout << jwb << endl;
  }
  return 0;
}