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;
vector<int> enumerate(int k) {
  vector<vector<int> > cur(k + 1);
  cur[0].push_back(0);
  cur[1].push_back(1);
  for (int i = 1; i < 26; i++) {
    vector<vector<int> > ncur = cur;
    for (int j = 1; j <= k && j <= i + 1; j++) {
      for (auto x : cur[j - 1]) {
        ncur[j].push_back(x ^ (1 << i));
      }
    }
    cur = ncur;
  }
  vector<int> ret;
  for (int i = 1; i <= k; i++) {
    for (int x : cur[i]) {
      ret.push_back(x);
    }
  }
  return ret;
}
const int N = 55;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
string t[N];
int n, m, k, px[N][N], py[N][N], dist[N][N];
inline void bfs(int x, int y) {
  memset(dist, -1, sizeof(dist));
  dist[x][y] = 0;
  queue<int> qx, qy;
  qx.push(x);
  qy.push(y);
  while (!qx.empty()) {
    x = qx.front();
    qx.pop();
    y = qy.front();
    qy.pop();
    for (int i = 0; i < 4; i++) {
      int nx = x + dx[i];
      int ny = y + dy[i];
      if (nx < 0 || nx >= n || ny < 0 || ny >= m || t[nx][ny] == '#') continue;
      if (dist[nx][ny] == -1) {
        dist[nx][ny] = dist[x][y] + 1;
        qx.push(nx);
        qy.push(ny);
      }
    }
  }
}
inline string getString(int sx, int sy, int fx, int fy) {
  string ret = "";
  vector<pair<int, int> > pos({{sx, sy}});
  int len = dist[sx][sy] - 1;
  for (int i = 0; i < len; i++) {
    vector<pair<int, int> > npos[26];
    for (auto pp : pos) {
      int x = pp.first;
      int y = pp.second;
      for (int j = 0; j < 4; j++) {
        int nx = x + dx[j];
        int ny = y + dy[j];
        if (nx < 0 || nx >= n || ny < 0 || ny >= m || t[nx][ny] == '#')
          continue;
        if (dist[nx][ny] + 1 == dist[x][y]) {
          npos[t[nx][ny] - 'a'].push_back({nx, ny});
        }
      }
    }
    for (int j = 0; j < 26; j++) {
      if (!npos[j].empty()) {
        ret += char('a' + j);
        pos = npos[j];
        break;
      }
    }
    sort(pos.begin(), pos.end());
    pos.resize(unique(pos.begin(), pos.end()) - pos.begin());
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(32);
  cin >> n >> m >> k;
  string s[n];
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  vector<int> masks = enumerate(k);
  string ans = "";
  bool flg = false;
  for (auto mask : masks) {
    int sx, sy, fx, fy;
    for (int i = 0; i < n; i++) {
      t[i] = s[i];
      for (int j = 0; j < m; j++) {
        if (t[i][j] == 'S') {
          sx = i;
          sy = j;
        } else if (t[i][j] == 'T') {
          fx = i;
          fy = j;
        } else {
          int ch = t[i][j] - 'a';
          if (((mask >> ch) & 1) == 0) t[i][j] = '#';
        }
      }
    }
    bfs(fx, fy);
    if (dist[sx][sy] == -1) continue;
    string cur = getString(sx, sy, fx, fy);
    bitset<26> bt(mask);
    if (flg) {
      if (ans.size() > cur.size())
        ans = cur;
      else if (ans.size() == cur.size())
        ans = min(ans, cur);
    } else {
      ans = cur;
      flg = true;
    }
  }
  if (flg)
    cout << ans;
  else
    cout << -1;
  return 0;
}