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;
int NLETT;
string letters;
set<string> seen;
string bestpath;
string paths[60][60];
int dist[60][60];
int d1[60][60], d2[60][60];
int visit[60][60];
void checkmin(string &a, const string &b) {
  if (a.length() > b.length()) a = b;
  if (a.length() == b.length() && a > b) a = b;
}
bool hasnext(vector<int> &alpha) {
  if (!NLETT) return false;
  int k = alpha.size();
  int p;
  for (p = k - 1; p >= 0; p--) {
    if (alpha[p] < NLETT + p - k) {
      alpha[p]++;
      break;
    }
  }
  if (p < 0) return false;
  for (++p; p < k; ++p) alpha[p] = alpha[p - 1] + 1;
  return true;
}
void bfs(pair<int, int> start, int n, int m, int usech[],
         const vector<string> &grid) {
  for (int r = 0; r < n; ++r)
    for (int c = 0; c < m; ++c) dist[r][c] = 2500;
  memset(visit, 0, sizeof visit);
  queue<pair<int, int> > que;
  que.push(start);
  visit[start.first][start.second] = 1;
  dist[start.first][start.second] = 0;
  while (!que.empty()) {
    pair<int, int> pos = que.front();
    que.pop();
    int r = pos.first, c = pos.second;
    int dir[][2] = {{-1, 0}, {0, +1}, {+1, 0}, {0, -1}};
    for (int x = 0; x < 4; ++x) {
      int rr = r + dir[x][0], cc = c + dir[x][1];
      if (0 <= rr && rr < n && 0 <= cc && cc < m && !visit[rr][cc]) {
        visit[rr][cc] = 1;
        dist[rr][cc] = dist[r][c] + 1;
        char ch = grid[rr][cc];
        if (('a' <= ch && ch <= 'z' && usech[ch - 'a'])) {
          que.push(pair<int, int>(rr, cc));
        }
      }
    }
  }
}
void updatepath(pair<int, int> start, pair<int, int> end,
                const vector<string> &grid, int usech[]) {
  int n = grid.size(), m = grid[0].size();
  int len = d1[end.first][end.second];
  memset(visit, 0, sizeof visit);
  queue<pair<int, int> > que;
  que.push(start);
  int r = start.first, c = start.second;
  visit[r][c] = 1;
  paths[r][c] = "";
  dist[r][c] = 0;
  while (!que.empty()) {
    pair<int, int> pos = que.front();
    que.pop();
    r = pos.first, c = pos.second;
    if (d1[r][c] >= len) break;
    string snew = paths[r][c] + grid[r][c];
    int dir[][2] = {{-1, 0}, {0, +1}, {+1, 0}, {0, -1}};
    for (int x = 0; x < 4; ++x) {
      int rr = r + dir[x][0], cc = c + dir[x][1];
      if (0 <= rr && rr < n && 0 <= cc && cc < m) {
        if (!visit[rr][cc]) {
          visit[rr][cc] = 1;
          paths[rr][cc] = snew;
          dist[rr][cc] = dist[r][c] + 1;
          char ch = grid[rr][cc];
          if ('a' <= ch && ch <= 'z' && usech[ch - 'a'])
            que.push(pair<int, int>(rr, cc));
        } else {
          checkmin(paths[rr][cc], snew);
        }
      }
    }
  }
  if (visit[end.first][end.second])
    checkmin(bestpath, paths[end.first][end.second]);
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<string> grid(n);
  for (int i = 0; i < n; ++i) cin >> grid[i];
  letters = "";
  int choice[26] = {0};
  pair<int, int> start, end;
  for (int r = 0; r < n; ++r)
    for (int c = 0; c < m; ++c) {
      if (grid[r][c] == 'S') start = pair<int, int>(r, c);
      if (grid[r][c] == 'T') end = pair<int, int>(r, c);
      if ('a' <= grid[r][c] && grid[r][c] <= 'z') choice[grid[r][c] - 'a'] = 1;
    }
  for (int i = 0; i < 26; ++i)
    if (choice[i]) letters += char(i + 'a');
  NLETT = letters.size();
  seen.clear();
  vector<int> alpha(k);
  for (int i = 0; i < k; ++i) alpha[i] = i;
  bestpath = string(2500, 'z');
  do {
    int usech[26] = {0};
    for (int i = 0; i < k; ++i) {
      char ch = letters[alpha[i]];
      usech[ch - 'a'] = 1;
    }
    bfs(start, n, m, usech, grid);
    memcpy(d1, dist, sizeof dist);
    updatepath(start, end, grid, usech);
  } while (hasnext(alpha));
  if (bestpath.length() >= 2500)
    cout << "-1" << endl;
  else
    cout << bestpath.substr(1) << endl;
}
int main() { solve(); }
