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;
const int N = 55;
int dist[N][N];
char allowed[4];
int n, m, maxAllowed;
int startx, starty, finx, finy;
string inf(N *N + 5, 'z' + 12);
string was[N][N];
string ans = inf;
string field[55];
int ansDist = N * N * N;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
inline bool isValidPoint(int x, int y) {
  return x >= 0 and x <= n - 1 and y >= 0 and y <= m - 1;
}
inline void upd(string &a, const string &b) {
  if (a.size() > b.size())
    a = b;
  else if (a.size() == b.size() and a > b)
    a = b;
}
void dfs(int x, int y) {
  if (was[x][y].size()) return;
  if (x == finx and y == finy) {
    was[x][y] = "";
    return;
  }
  string &ret = was[x][y];
  ret = inf;
  for (int i = 0; i < 4; i++) {
    int nx = x + dx[i];
    int ny = y + dy[i];
    if (not isValidPoint(nx, ny)) continue;
    bool isValid = false;
    for (int j = 0; j < maxAllowed; j++)
      if (allowed[j] == field[nx][ny]) isValid = true;
    if (nx == finx and ny == finy) isValid = true;
    if (not isValid) continue;
    if (dist[nx][ny] + 1 != dist[x][y]) continue;
    dfs(nx, ny);
    upd(ret, was[nx][ny]);
  }
  ret = field[x][y] + ret;
}
string recover() {
  string ret = "";
  set<pair<int, int> > cur;
  cur.insert(make_pair(startx, starty));
  if (dist[startx][starty] == 0) return inf;
  while (not cur.empty()) {
    set<pair<int, int> > next;
    char minLetter = 'z' + 1;
    for (auto &t : cur) {
      for (int i = 0; i < 4; i++) {
        int nx = t.first + dx[i];
        int ny = t.second + dy[i];
        if (not isValidPoint(nx, ny)) continue;
        if (dist[nx][ny] + 1 != dist[t.first][t.second]) continue;
        if (nx == finx and ny == finy)
          minLetter = 'a' - 23;
        else {
          minLetter = min(minLetter, field[nx][ny]);
        }
      }
    }
    if (minLetter >= 'a') ret += minLetter;
    for (auto &t : cur) {
      for (int i = 0; i < 4; i++) {
        int nx = t.first + dx[i];
        int ny = t.second + dy[i];
        if (not isValidPoint(nx, ny)) continue;
        if (dist[nx][ny] + 1 != dist[t.first][t.second]) continue;
        if (field[nx][ny] == minLetter) next.insert(make_pair(nx, ny));
      }
    }
    cur = next;
  }
  return ret;
}
void solve() {
  memset(dist, 0, sizeof(dist));
  queue<pair<int, int> > q;
  q.push(make_pair(finx, finy));
  while (not q.empty()) {
    pair<int, int> cur = q.front();
    q.pop();
    int cx = cur.first, cy = cur.second;
    for (int i = 0; i < 4; i++) {
      int nx = cx + dx[i];
      int ny = cy + dy[i];
      if (nx == finx and ny == finy) continue;
      if (not isValidPoint(nx, ny)) continue;
      if (dist[nx][ny] > 0) continue;
      bool isValid = false;
      if (nx != startx or ny != starty) {
        for (int j = 0; j < maxAllowed; j++)
          if (allowed[j] == field[nx][ny]) isValid = true;
      } else
        isValid = true;
      if (not isValid) continue;
      dist[nx][ny] = dist[cx][cy] + 1;
      q.push(make_pair(nx, ny));
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) was[i][j] = "";
  if (dist[startx][starty] == 0 or dist[startx][starty] > ansDist) return;
  ansDist = dist[startx][starty];
  upd(ans, recover());
}
void rec(int cur) {
  if (cur == maxAllowed) {
    bool isValid = false;
    for (int i = 0; i < 4; i++) {
      int nx = startx + dx[i];
      int ny = starty + dy[i];
      if (nx == finx and ny == finy) isValid = true;
      if (not isValidPoint(nx, ny)) continue;
      bool isPresent = false;
      for (int j = 0; j < maxAllowed; j++) {
        if (field[nx][ny] == allowed[j]) isPresent = true;
      }
      if (isPresent) isValid = true;
    }
    if (isValid) solve();
  } else {
    for (int i = (cur == 0 ? 0 : allowed[cur - 1] - 'a' + 1); i < 26; i++) {
      bool isValid = true;
      for (int j = 0; j < cur; j++)
        if (allowed[j] == i + 'a') isValid = false;
      if (isValid) {
        allowed[cur] = i + 'a';
        rec(cur + 1);
      }
    }
  }
}
int main() {
  cin >> n >> m >> maxAllowed;
  for (int i = 0; i < n; i++) {
    cin >> field[i];
    for (int j = 0; j < m; j++)
      if (field[i][j] == 'S') {
        startx = i;
        starty = j;
      } else if (field[i][j] == 'T') {
        finx = i;
        finy = j;
      }
  }
  rec(0);
  if (ans.size() and ans[0] == 'S') ans = ans.substr(1);
  if (ans == inf)
    cout << -1;
  else
    cout << ans << endl;
  return 0;
}