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;
char M[52][52];
int P[26], dx[] = {0, 1, -1, 0}, dy[] = {1, 0, 0, -1};
int main() {
  int X, Y, K;
  scanf("%d%d%d", &X, &Y, &K);
  for (int i = 0; i < K; i++) P[i] = 1;
  sort(P, P + 26);
  for (int i = 0; i < X; i++) scanf("%s", M[i]);
  pair<int, int> S, T;
  for (int i = 0; i < X; i++)
    for (int j = 0; j < Y; j++)
      if (M[i][j] == 'S')
        S = pair<int, int>(i, j);
      else if (M[i][j] == 'T')
        T = pair<int, int>(i, j);
  string result = "-1";
  do {
    queue<pair<int, int> > q;
    vector<vector<int> > d(X, vector<int>(Y, -1));
    q.push(T);
    d[T.first][T.second] = 0;
    while (!q.empty()) {
      pair<int, int> p = q.front();
      q.pop();
      if (p == S) break;
      for (int i = 0; i < 4; i++) {
        int x = p.first + dx[i], y = p.second + dy[i];
        if (0 <= x && x < X && 0 <= y && y < Y && d[x][y] == -1 &&
            (M[x][y] == 'S' || P[M[x][y] - 'a'])) {
          q.push(pair<int, int>(x, y));
          d[x][y] = d[p.first][p.second] + 1;
        }
      }
    }
    int distance = d[S.first][S.second];
    if (distance == -1) continue;
    string r = "";
    vector<pair<int, int> > v, nv;
    v.push_back(S);
    while (distance > 1) {
      char c = 'z';
      for (size_t i = 0; i < v.size(); i++) {
        for (int j = 0; j < 4; j++) {
          int x = v[i].first + dx[j], y = v[i].second + dy[j];
          if (0 <= x && x < X && 0 <= y && y < Y && d[x][y] == distance - 1)
            c = min(c, M[x][y]);
        }
      }
      for (size_t i = 0; i < v.size(); i++) {
        for (int j = 0; j < 4; j++) {
          int x = v[i].first + dx[j], y = v[i].second + dy[j];
          if (0 <= x && x < X && 0 <= y && y < Y && d[x][y] == distance - 1 &&
              M[x][y] == c)
            nv.push_back(pair<int, int>(x, y));
        }
      }
      sort(nv.begin(), nv.end());
      nv.resize(unique(nv.begin(), nv.end()) - nv.begin());
      v = nv;
      nv.clear();
      r += c;
      distance--;
    }
    if (result == "-1" || r.length() < result.length() ||
        (r.length() == result.length() && r < result))
      result = r;
  } while (next_permutation(P, P + 26));
  printf("%s\n", result.c_str());
  return 0;
}