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 n, m, k;
char mapa[55][55];
pair<int, int> ini, fim;
int dX[] = {1, 0, -1, 0};
int dY[] = {0, -1, 0, 1};
set<int> set_cores[55][55];
struct evento {
  int x, y, cores, dist;
  string caminho;
  evento(int x, int y) : x(x), y(y), cores(0), dist(0), caminho("") {}
  evento(int x, int y, int d, int co, string c)
      : x(x), y(y), cores(co), dist(d), caminho(c) {}
  bool operator<(const evento& o) const {
    return tie(dist, caminho) > tie(o.dist, o.caminho);
  }
};
bool invalid(int nx, int ny) { return nx < 0 || nx >= n || ny < 0 || ny >= m; }
void dijkstra() {
  priority_queue<evento> pfila;
  pfila.push(evento(ini.first, ini.second));
  while (!pfila.empty()) {
    evento u = pfila.top();
    pfila.pop();
    int x = u.x, y = u.y;
    if (mapa[x][y] == 'T') {
      cout << u.caminho << endl;
      return;
    }
    for (int i = 0; i < 4; i++) {
      int nx = x + dX[i];
      int ny = y + dY[i];
      string ncaminho = u.caminho;
      int ncores = u.cores;
      if (invalid(nx, ny)) continue;
      if (mapa[nx][ny] >= 'a' && mapa[nx][ny] <= 'z') {
        ncaminho += mapa[nx][ny];
        ncores |= (1 << (mapa[nx][ny] - 'a'));
      }
      int flag_visitado = 0;
      for (auto cor : set_cores[nx][ny]) {
        if ((cor | ncores) == ncores) {
          flag_visitado = 1;
          break;
        }
      }
      if (flag_visitado) continue;
      if (__builtin_popcount(ncores) <= k) {
        set_cores[nx][ny].insert(ncores);
        pfila.push(evento(nx, ny, u.dist + 1, ncores, ncaminho));
      }
    }
  }
  cout << "-1" << endl;
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", mapa[i]);
    for (int j = 0; j < m; j++) {
      if (mapa[i][j] == 'S') {
        ini = {i, j};
      }
      if (mapa[i][j] == 'T') {
        fim = {i, j};
      }
    }
  }
  dijkstra();
  return 0;
}
