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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m, k;
int si, sj, ti, tj;
int was[50][50];
string g[50], res[50][50];
pair<int, int> q[2500];
int x[50][50];
int can[27];
string ANS;
int ans;
int go() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      was[i][j] = 0;
    }
  can[26] = 1;
  int l = 0, r = 1;
  q[0] = make_pair(si, sj);
  was[si][sj] = 1;
  while (l < r) {
    int i = q[l].first;
    int j = q[l].second;
    if (i == ti && j == tj) break;
    l++;
    for (int t = 0; t < 4; t++) {
      int ni = i + int(t == 0) - int(t == 1);
      int nj = j + int(t == 2) - int(t == 3);
      if (ni >= 0 && nj >= 0 && ni < n && nj < m && can[x[ni][nj]]) {
        if (!was[ni][nj]) {
          was[ni][nj] = was[i][j] + 1;
          q[r++] = make_pair(ni, nj);
        }
      }
    }
  }
  if (!was[ti][tj] || was[ti][tj] > ans) return 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) res[i][j] = "";
  was[si][sj] = 1;
  res[si][sj] = "S";
  for (int l = 0; l < r; l++) {
    int i = q[l].first;
    int j = q[l].second;
    if (i == ti && j == tj) break;
    for (int t = 0; t < 4; t++) {
      int ni = i + int(t == 0) - int(t == 1);
      int nj = j + int(t == 2) - int(t == 3);
      if (ni >= 0 && nj >= 0 && ni < n && nj < m && can[x[ni][nj]] &&
          was[ni][nj] == was[i][j] + 1) {
        string tmp = res[i][j] + g[ni][nj];
        if (res[ni][nj] == "" || res[ni][nj] > tmp) res[ni][nj] = tmp;
      }
    }
  }
  if (was[ti][tj] < ans) {
    ans = was[ti][tj];
    ANS = res[ti][tj];
  } else if (was[ti][tj] == ans)
    ANS = min(ANS, res[ti][tj]);
  return 0;
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> g[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (g[i][j] == 'S') {
        x[i][j] = 26;
        si = i;
        sj = j;
      } else if (g[i][j] == 'T') {
        x[i][j] = 26;
        ti = i;
        tj = j;
      } else
        x[i][j] = g[i][j] - 'a';
  ans = 1e9;
  ANS = "";
  for (int a = 0; a < 26; a++) {
    can[a] = 1;
    if (k == 1)
      go();
    else
      for (int b = a + 1; b < 26; b++) {
        can[b] = 1;
        if (k == 2)
          go();
        else
          for (int c = b + 1; c < 26; c++) {
            can[c] = 1;
            if (k == 3)
              go();
            else
              for (int d = c + 1; d < 26; d++) {
                can[d] = 1;
                go();
                can[d] = 0;
              }
            can[c] = 0;
          }
        can[b] = 0;
      }
    can[a] = 0;
  }
  if (ANS == "")
    printf("-1\n");
  else
    printf("%s\n", ANS.substr(1, ((int)(ANS).size()) - 2).c_str());
  return 0;
}
