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 R, C, K, Tail;
int Board[60][60], Road[60][60], Best[60][60], Oki[60][60];
int Allow[100000][5];
int Last[60][60];
char Now[5];
int Res;
int Check[30];
char Ans[10000];
int QR[10000], QC[10000];
void pre_process(char now, int left) {
  if (now > 'z') {
    if (left == 0) {
      int i;
      for (i = 0; i < (int)(K); i++) Allow[Tail][i] = (int)(Now[i] - 'a');
      Tail++;
    }
    return;
  }
  pre_process(now + 1, left);
  if (left > 0) {
    Now[left - 1] = now;
    pre_process(now + 1, left - 1);
  }
}
bool isValid(int r, int c) {
  return 0 <= r && r < R && 0 <= c && c < C && Road[r][c];
}
int main() {
  int i, j, k, r, c;
  int sr, sc, er, ec;
  char a;
  int dr[4] = {-1, 0, 1, 0};
  int dc[4] = {0, 1, 0, -1};
  int tt, et;
  scanf("%d%d%d", &R, &C, &K);
  for (i = 0; i < (int)(R); i++) {
    for (j = 0; j < (int)(C); j++) {
      scanf("%c", &a);
      if (!('a' <= a && a <= 'z') && a != 'S' && a != 'T') {
        j--;
        continue;
      }
      if ('a' <= a && a <= 'z')
        Board[i][j] = a - 'a';
      else if (a == 'S')
        Board[i][j] = -1, sr = i, sc = j;
      else
        Board[i][j] = -2, er = i, ec = j;
    }
  }
  Tail = 0;
  pre_process('a', K);
  Res = -1;
  for (i = 0; i < (int)(Tail); i++) {
    for (j = 0; j < (int)(26); j++) Check[j] = 0;
    for (j = 0; j < (int)(K); j++) Check[Allow[i][j]] = 1;
    for (r = 0; r < (int)(R); r++)
      for (c = 0; c < (int)(C); c++) {
        if (Board[r][c] >= 0)
          Road[r][c] = Check[Board[r][c]];
        else
          Road[r][c] = 1;
        Best[r][c] = -1;
        Oki[r][c] = 0;
        Last[r][c] = 0;
      }
    tt = 1;
    QR[0] = sr, QC[0] = sc;
    Best[sr][sc] = 0;
    for (j = 0; j < (int)(tt); j++) {
      r = QR[j], c = QC[j];
      for (k = 0; k < (int)(4); k++)
        if (isValid(r + dr[k], c + dc[k]) && Best[r + dr[k]][c + dc[k]] == -1) {
          Best[r + dr[k]][c + dc[k]] = Best[r][c] + 1;
          QR[tt] = r + dr[k], QC[tt] = c + dc[k];
          tt++;
        }
    }
    if (Best[er][ec] == -1 || (Res != -1 && Best[er][ec] > Res)) continue;
    bool update = false;
    if (Res == -1 || Best[er][ec] < Res) update = true;
    Res = Best[er][ec];
    Ans[Res - 1] = Ans[Res] = 0;
    tt = 1;
    QR[0] = er, QC[0] = ec;
    Oki[er][ec] = 1;
    for (j = 0; j < (int)(tt); j++) {
      r = QR[j], c = QC[j];
      for (k = 0; k < (int)(4); k++)
        if (isValid(r + dr[k], c + dc[k]) &&
            Best[r + dr[k]][c + dc[k]] == Best[r][c] - 1 &&
            !Oki[r + dr[k]][c + dc[k]]) {
          Oki[r + dr[k]][c + dc[k]] = 1;
          QR[tt] = r + dr[k], QC[tt] = c + dc[k];
          tt++;
        }
    }
    int cur = 0, tt = 1, move = 0;
    QR[0] = sr, QC[0] = sc;
    Best[sr][sc] = 0;
    while (cur < Res - 1) {
      et = tt;
      int best = -1;
      for (j = move; j < et; j++) {
        r = QR[j], c = QC[j];
        for (k = 0; k < (int)(4); k++)
          if (isValid(r + dr[k], c + dc[k]) && Oki[r + dr[k]][c + dc[k]] &&
              Best[r + dr[k]][c + dc[k]] == cur + 1) {
            if (best == -1 || best > Board[r + dr[k]][c + dc[k]]) {
              best = Board[r + dr[k]][c + dc[k]];
            }
          }
      }
      if (!update && (char)(best + 'a') > Ans[cur]) break;
      if (!update && (char)(best + 'a') < Ans[cur]) update = true;
      Ans[cur] = (char)(best + 'a');
      cur++;
      for (j = move; j < et; j++) {
        r = QR[j], c = QC[j];
        for (k = 0; k < (int)(4); k++)
          if (isValid(r + dr[k], c + dc[k]) && Oki[r + dr[k]][c + dc[k]] &&
              Best[r + dr[k]][c + dc[k]] == cur &&
              !Last[r + dr[k]][c + dc[k]]) {
            if (best == Board[r + dr[k]][c + dc[k]]) {
              Last[r + dr[k]][c + dc[k]] = 1;
              QR[tt] = r + dr[k], QC[tt] = c + dc[k];
              tt++;
            }
          }
      }
      move = et;
    }
  }
  if (Res == -1)
    printf("%d\n", Res);
  else
    printf("%s\n", Ans);
  return 0;
}
