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>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
string filename = "test";
int n, m, k;
vector<string> mapa;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
bool allowed[26];
int ctoi(char c) { return c - 'a'; }
int valid(int i, int j) { return i >= 0 && i < n && j >= 0 && j < m; }
pair<int, int> Q[100000];
int D[51][51];
int main() {
  scanf("%d %d %d", &n, &m, &k);
  string tmp_str;
  for (int _n(n), i(0); i < _n; i++) {
    cin >> tmp_str;
    mapa.push_back(tmp_str);
  }
  pair<int, int> start, end;
  vector<char> all_letters;
  for (int _n(n), i(0); i < _n; i++)
    for (int _n(m), j(0); j < _n; j++) {
      if (mapa[i][j] == 'S') {
        start = pair<int, int>(i, j);
      } else if (mapa[i][j] == 'T') {
        end = pair<int, int>(i, j);
      } else {
        all_letters.push_back(mapa[i][j]);
      }
    }
  for (int _n(4), dir(0); dir < _n; dir++) {
    int nx = start.first + dx[dir], ny = start.second + dy[dir];
    if (valid(nx, ny) && pair<int, int>(nx, ny) == end) {
      cout << "\n";
      return 0;
    }
  }
  sort(((all_letters)).begin(), ((all_letters)).end()),
      (all_letters)
          .erase(unique(((all_letters)).begin(), ((all_letters)).end()),
                 (all_letters).end());
  int nn = all_letters.size();
  int ires = -1;
  string sres = "";
  k = (((k) < (nn)) ? (k) : (nn));
  unsigned int mask = (1 << k) - 1;
  while (!(mask & (1 << nn))) {
    memset(allowed, false, sizeof(allowed));
    for (int _n(nn), i(0); i < _n; i++)
      if (mask & (1 << i)) {
        allowed[ctoi(all_letters[i])] = true;
      }
    int wr = 0, rd = 0;
    memset(D, -1, sizeof(D));
    Q[wr++] = end;
    D[end.first][end.second] = 0;
    while (rd < wr) {
      pair<int, int> now = Q[rd++];
      for (int _n(4), dir(0); dir < _n; dir++) {
        int nx = now.first + dx[dir], ny = now.second + dy[dir];
        if (!valid(nx, ny)) continue;
        if (D[nx][ny] != -1) continue;
        if (mapa[nx][ny] != 'T' && mapa[nx][ny] != 'S' &&
            !allowed[ctoi(mapa[nx][ny])])
          continue;
        D[nx][ny] = D[now.first][now.second] + 1;
        Q[wr++] = pair<int, int>(nx, ny);
      }
    }
    int ires_now = D[start.first][start.second];
    if (ires_now != -1 && !(ires != -1 && ires_now > ires)) {
      string sres_now = "";
      vector<pair<int, int> > pos[2];
      int cur_index = 0;
      pos[cur_index].push_back(start);
      int d = D[start.first][start.second];
      while (d != 0) {
        sres_now.push_back(
            mapa[pos[cur_index][0].first][pos[cur_index][0].second]);
        int next_index = cur_index ^ 1;
        bool exit_found = false;
        char min_c = (char)('z' + 1);
        for (int _n(pos[cur_index].size()), i(0); i < _n; i++) {
          for (int _n(4), dir(0); dir < _n; dir++) {
            int nx = pos[cur_index][i].first + dx[dir],
                ny = pos[cur_index][i].second + dy[dir];
            if (!valid(nx, ny)) continue;
            if (D[nx][ny] == -1) continue;
            if (D[nx][ny] + 1 != d) continue;
            if (mapa[nx][ny] == 'T') {
              exit_found = true;
              break;
            }
            D[nx][ny] = -1;
            if (mapa[nx][ny] < min_c) {
              min_c = mapa[nx][ny];
              pos[next_index].clear();
              pos[next_index].push_back(pair<int, int>(nx, ny));
            } else if (mapa[nx][ny] == min_c) {
              pos[next_index].push_back(pair<int, int>(nx, ny));
            }
          }
          if (exit_found) break;
        }
        if (exit_found) break;
        --d;
        cur_index = next_index;
      }
      if (ires == -1 || ires_now < ires) {
        ires = ires_now;
        sres = sres_now;
      } else if (ires_now == ires && sres_now < sres) {
        ires = ires_now;
        sres = sres_now;
      }
    }
    int lo = mask & ~(mask - 1);
    int lz = (mask + lo) & ~mask;
    mask |= lz;
    mask &= ~(lz - 1);
    mask |= (lz / lo / 2) - 1;
  }
  if (ires == -1) {
    printf("-1\n");
  } else {
    cout << sres.substr(1) << endl;
  }
  return 0;
}
