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;
inline int GetBit(int mask, int num) { return (mask >> num) & 1; }
inline int GetAmountOfBits(int mask) {
  int ans = 0;
  while (mask > 0) {
    ans += (mask & 1);
    mask >>= 1;
  }
  return ans;
}
const int kAlphabetSize = 26;
const int kDx[4] = {-1, 0, 0, 1};
const int kDy[4] = {0, -1, 1, 0};
int n, m, k;
vector<vector<char> > field;
int si, sj;
int fi, fj;
bool ans_found;
int ans_len;
string ans_path;
vector<vector<bool> > used;
vector<vector<bool> > inused;
vector<vector<int> > dist;
vector<int> char_posses_i[kAlphabetSize];
vector<int> char_posses_j[kAlphabetSize];
void ClearUsed(int mask, vector<vector<bool> >& used) {
  for (int c = 0; c < kAlphabetSize; ++c) {
    if (!GetBit(mask, c)) {
      continue;
    }
    for (int i = 0; i < (int)char_posses_i[c].size(); ++i) {
      used[char_posses_i[c][i]][char_posses_j[c][i]] = false;
    }
  }
  used[si][sj] = false;
  used[fi][fj] = false;
}
void UpdateAns(int mask) {
  if (dist[fi][fj] == 1) {
    ans_found = true;
    ans_len = 1;
    ans_path = "";
    return;
  }
  string cur_path;
  cur_path.reserve(dist[fi][fj] - 1);
  deque<int> qi;
  deque<int> qj;
  qi.push_back(fi);
  qj.push_back(fj);
  inused[fi][fj] = true;
  while (!qi.empty()) {
    int ci = qi.front();
    qi.pop_front();
    int cj = qj.front();
    qj.pop_front();
    for (int step_index = 0; step_index < 4; ++step_index) {
      int ni = ci + kDx[step_index];
      int nj = cj + kDy[step_index];
      if (ni >= 0 && ni < n && nj >= 0 && nj < m && !inused[ni][nj] &&
          used[ni][nj] && dist[ni][nj] + 1 == dist[ci][cj]) {
        inused[ni][nj] = true;
        qi.push_back(ni);
        qj.push_back(nj);
      }
    }
  }
  qi.push_back(si);
  qj.push_back(sj);
  used[si][sj] = false;
  while (!qi.empty()) {
    char min_char = field[qi[0]][qj[0]];
    for (int i = 1; i < (int)qi.size(); ++i) {
      min_char = min(min_char, field[qi[i]][qj[i]]);
    }
    if (min_char != 'S' && min_char != 'T') {
      cur_path.push_back(min_char);
    }
    int sz = qi.size();
    for (int itr = 0; itr < (int)sz; ++itr) {
      int ci = qi.front();
      qi.pop_front();
      int cj = qj.front();
      qj.pop_front();
      if (field[ci][cj] != min_char) {
        continue;
      }
      for (int step_index = 0; step_index < 4; ++step_index) {
        int ni = ci + kDx[step_index];
        int nj = cj + kDy[step_index];
        if (ni >= 0 && ni < n && nj >= 0 && nj < m && inused[ni][nj] &&
            used[ni][nj] && dist[ni][nj] == dist[ci][cj] + 1) {
          used[ni][nj] = false;
          qi.push_back(ni);
          qj.push_back(nj);
        }
      }
    }
  }
  ClearUsed(mask, inused);
  if (!ans_found || dist[fi][fj] < ans_len || cur_path < ans_path) {
    ans_found = true;
    ans_len = dist[fi][fj];
    ans_path = cur_path;
  }
}
void Solve(int mask) {
  queue<int> qi;
  queue<int> qj;
  used[si][sj] = true;
  dist[si][sj] = 0;
  qi.push(si);
  qj.push(sj);
  while (!qi.empty()) {
    int ci = qi.front();
    qi.pop();
    int cj = qj.front();
    qj.pop();
    for (int step_index = 0; step_index < 4; ++step_index) {
      int ni = ci + kDx[step_index];
      int nj = cj + kDy[step_index];
      if (ni >= 0 && ni < n && nj >= 0 && nj < m &&
          (field[ni][nj] == 'T' || field[ni][nj] == 'S' ||
           GetBit(mask, field[ni][nj] - 'a')) &&
          !used[ni][nj]) {
        used[ni][nj] = true;
        dist[ni][nj] = dist[ci][cj] + 1;
        qi.push(ni);
        qj.push(nj);
      }
    }
  }
  if (used[fi][fj]) {
    if (!ans_found || dist[fi][fj] <= ans_len) {
      UpdateAns(mask);
    }
  }
  ClearUsed(mask, used);
}
void Dfs(int c, int mask, int num) {
  if (c == kAlphabetSize) {
    Solve(mask);
    return;
  }
  if (num + 1 <= k) {
    Dfs(c + 1, mask + (1 << c), num + 1);
  }
  Dfs(c + 1, mask, num);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> k;
  field.resize(n, vector<char>(m));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> field[i][j];
      if (field[i][j] == 'S') {
        si = i;
        sj = j;
      }
      if (field[i][j] == 'T') {
        fi = i;
        fj = j;
      }
    }
  }
  used.assign(n, vector<bool>(m, false));
  inused.assign(n, vector<bool>(m, false));
  dist.resize(n, vector<int>(m));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (field[i][j] >= 'a' && field[i][j] <= 'z') {
        char_posses_i[field[i][j] - 'a'].push_back(i);
        char_posses_j[field[i][j] - 'a'].push_back(j);
      }
    }
  }
  ans_found = false;
  Dfs(0, 0, 0);
  if (!ans_found) {
    cout << -1 << endl;
  } else {
    cout << ans_path << endl;
  }
  return 0;
}