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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = 1000 * 1000 * 1000;
char tbl[55][55];
bool ap[55][55];
int dst[55][55];
string p[55][55];
bool got[55];
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, -1, 0, 1};
string path;
bool SUKASUKA;
int N, M, K;
pair<int, int> s, t;
bool checkPos(int x, int y) {
  return x >= 0 && x < N && y >= 0 && y < M && ap[x][y];
}
void solve() {
  if (path.empty()) {
    path.resize(N * M + 1, 'z' + 1);
  }
  vector<char> gc;
  for (int i = (0); i < (26); ++i)
    if (got[i]) {
      gc.push_back(char('a' + i));
    }
  for (int i = (0); i < (N); ++i)
    for (int j = (0); j < (M); ++j) {
      ap[i][j] = false;
      for (int k = (0); k < ((int)gc.size()); ++k)
        if (tbl[i][j] == gc[k]) {
          ap[i][j] = true;
        }
    }
  ap[s.first][s.second] = ap[t.first][t.second] = true;
  for (int i = (0); i < (N); ++i)
    for (int j = (0); j < (M); ++j) {
      dst[i][j] = INF;
      p[i][j] = "";
    }
  dst[s.first][s.second] = 0;
  queue<pair<int, int> > q;
  q.push(s);
  while (!q.empty()) {
    pair<int, int> now = q.front();
    q.pop();
    for (int k = (0); k < (4); ++k) {
      int x = now.first + dx[k], y = now.second + dy[k];
      if (checkPos(x, y))
        if (dst[x][y] > dst[now.first][now.second] + 1) {
          dst[x][y] = dst[now.first][now.second] + 1;
          p[x][y] = p[now.first][now.second] + tbl[x][y];
          q.push(make_pair(x, y));
        } else if (dst[x][y] == dst[now.first][now.second] + 1) {
          p[x][y] = min(p[x][y], p[now.first][now.second] + tbl[x][y]);
        }
    }
  }
  if (dst[t.first][t.second] == INF) {
    return;
  }
  SUKASUKA = true;
  string npath = p[t.first][t.second];
  if ((int)npath.size() < (int)path.size() ||
      (int)npath.size() == (int)path.size() && npath < path) {
    path = npath;
  }
}
void gen(int step = 0, int prev = -1) {
  if (step == K) {
    solve();
  } else {
    if (prev + 1 == 26) {
      return;
    }
    for (int i = (prev + 1); i < (26); ++i) {
      got[i] = true;
      gen(step + 1, i);
      got[i] = false;
    }
  }
}
void read() {
  cin >> N >> M >> K;
  for (int i = (0); i < (N); ++i)
    for (int j = (0); j < (M); ++j) {
      cin >> tbl[i][j];
      if (tbl[i][j] == 'S') {
        s.first = i;
        s.second = j;
      }
      if (tbl[i][j] == 'T') {
        t.first = i;
        t.second = j;
      }
    }
}
void out() {
  if (SUKASUKA) {
    for (int i = (0); i < ((int)path.size() - 1); ++i) {
      cout << path[i];
    }
  } else {
    cout << "-1";
  }
}
int main() {
  read();
  gen();
  out();
  return 0;
}
