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;
const int MAXN = 55, dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
int n, m, k, dis[MAXN][MAXN], sx, sy, tx, ty;
queue<pair<int, int> > S;
char mp[MAXN][MAXN];
string ans, res;
bool r[27], o;
bool isval(int x, int y) { return x < n && y < m && x >= 0 && y >= 0; }
void BFS(vector<pair<int, int> > w) {
  vector<pair<int, int> > rx[27];
  for (int ii = 0; ii < w.size(); ii++) {
    int x = w[ii].first, y = w[ii].second;
    for (int i = 0; i < 4; i++) {
      int nx = x + dx[i], ny = y + dy[i];
      if (!isval(nx, ny) || (!r[mp[nx][ny] - 'a'] && mp[nx][ny] != 'T') ||
          dis[nx][ny] + 1 != dis[x][y])
        continue;
      if (mp[nx][ny] == 'T') return;
      rx[mp[nx][ny] - 'a'].push_back({nx, ny});
    }
  }
  for (int i = 0; i < 27; i++) {
    sort(rx[i].begin(), rx[i].end());
    rx[i].resize(unique(rx[i].begin(), rx[i].end()) - rx[i].begin());
    if (rx[i].size() == 0) continue;
    BFS(rx[i]);
    res += (char)('a' + i);
    return;
  }
}
void solve() {
  res.clear();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) dis[i][j] = -1;
  S.push({tx, ty});
  dis[tx][ty] = 0;
  while (!S.empty()) {
    pair<int, int> c = S.front();
    S.pop();
    for (int i = 0; i < 4; i++) {
      int nx = c.first + dx[i], ny = c.second + dy[i];
      if (!isval(nx, ny) ||
          (mp[nx][ny] != 'S' && mp[nx][ny] != 'T' && !r[mp[nx][ny] - 'a']))
        continue;
      if (dis[nx][ny] == -1) {
        S.push({nx, ny});
        dis[nx][ny] = dis[c.first][c.second] + 1;
      }
    }
  }
  if (dis[sx][sy] == -1) return;
  vector<pair<int, int> > vc;
  vc.push_back({sx, sy});
  BFS(vc);
  reverse(res.begin(), res.end());
  if (!o)
    ans = res, o = 1;
  else {
    if (ans.size() < res.size()) return;
    if (ans.size() > res.size())
      ans = res;
    else
      ans = min(ans, res);
  }
}
void bt(int ind, int rem) {
  if (ind == 26) {
    solve();
    return;
  }
  if (rem > 0) {
    r[ind] = 1;
    bt(ind + 1, rem - 1);
  }
  r[ind] = 0;
  bt(ind + 1, rem);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> mp[i][j];
      if (mp[i][j] == 'S')
        sx = i, sy = j;
      else if (mp[i][j] == 'T')
        tx = i, ty = j;
    }
  bt(0, k);
  if (!o)
    cout << -1 << endl;
  else
    cout << ans << endl;
}
