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 dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int N, M, K;
int F[60][60], Q[3000][2];
int SX, SY, TX, TY;
string P[60][60];
pair<char, int> A[4];
string Ret;
char Map[60][60];
bool G[60][60], Choose[30];
inline void BFS() {
  memset(G, false, sizeof(G));
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++)
      if (Choose[Map[i][j] - 'a']) {
        G[i][j] = true;
      }
  }
  G[SX][SY] = true;
  G[TX][TY] = true;
  memset(F, -1, sizeof(F));
  F[SX][SY] = 0;
  int Head = 0, Tail = 1;
  Q[1][0] = SX;
  Q[1][1] = SY;
  while (Head++ < Tail) {
    int x = Q[Head][0];
    int y = Q[Head][1];
    if (x == TX && y == TY) {
      break;
    }
    if (x != SX || y != SY) {
      P[x][y] += Map[x][y];
    }
    for (int i = 0; i < 4; i++) {
      int u = x + dx[i];
      int v = y + dy[i];
      if (!G[u][v]) {
        continue;
      }
      if (F[u][v] == -1) {
        F[u][v] = F[x][y] + 1;
        P[u][v] = P[x][y];
        Tail++;
        Q[Tail][0] = u;
        Q[Tail][1] = v;
      } else if (F[x][y] + 1 == F[u][v] && P[x][y] < P[u][v]) {
        P[u][v] = P[x][y];
      }
    }
  }
  if (F[TX][TY] == -1) {
    return;
  }
  if (Ret != "-1" && F[TX][TY] > Ret.size() + 1) {
    return;
  }
  string Ans = P[TX][TY];
  if (Ret == "-1" || Ans.size() < Ret.size() || Ans < Ret) {
    Ret = Ans;
  }
}
inline void Search(int k, int total) {
  if (total > K) {
    return;
  }
  if (k == 26) {
    if (total == K) {
      BFS();
    }
  } else {
    Choose[k] = true;
    Search(k + 1, total + 1);
    Choose[k] = false;
    Search(k + 1, total);
  }
}
int main(int argc, char *argv[]) {
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%s", Map[i] + 1);
    for (int j = 1; j <= M; j++) {
      if (Map[i][j] == 'S') {
        SX = i;
        SY = j;
        Map[i][j] = 'a';
      }
      if (Map[i][j] == 'T') {
        TX = i;
        TY = j;
        Map[i][j] = 'a';
      }
    }
  }
  Ret = "-1";
  Search(0, 0);
  cout << Ret << endl;
  return 0;
}
