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>
#pragma warning(disable : 4996)
using namespace std;
const int MAXN = 55;
struct Node {
  int sta, pos;
  string path;
  Node(string _path, int _sta, int _pos) : path(_path), sta(_sta), pos(_pos) {}
  bool operator<(const Node& v) const {
    int len = path.size();
    int lenb = v.path.size();
    if (len != lenb) return len > lenb;
    if (path < v.path) return false;
    if (path > v.path) return true;
    return pos < v.pos;
  }
};
int n, m, K;
int si, sj;
static int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
char mat[MAXN][MAXN];
int visited[MAXN][MAXN];
set<long long> s;
void InputMat() {
  for (int i = 0; i < n; i++) {
    while (getchar() != '\n')
      ;
    for (int j = 0; j < m; j++) {
      mat[i][j] = getchar();
      if (mat[i][j] == 'S') {
        si = i;
        sj = j;
      }
    }
  }
}
int iCount(int x) {
  int ret = 0;
  for (; x; x &= x - 1) ret++;
  return ret;
}
bool legal(int x, int y) {
  return x >= 0 && x < n && y >= 0 && y < m && mat[x][y] != 'S';
}
string Bfs() {
  priority_queue<Node> q;
  memset(visited, 0, sizeof(visited));
  s.clear();
  q.push(Node("", 0, si * m + sj));
  s.insert(si * m + sj);
  while (!q.empty()) {
    Node u = q.top();
    q.pop();
    for (int k = 0; k < 4; k++) {
      int nx = u.pos / m + dir[k][0];
      int ny = u.pos % m + dir[k][1];
      if (legal(nx, ny) == false) continue;
      if (mat[nx][ny] == 'T') {
        return u.path;
      }
      if (!(u.sta & (1 << (mat[nx][ny] - 'a'))) && iCount(u.sta) == K) continue;
      Node v = Node(u.path + mat[nx][ny], u.sta | (1 << (mat[nx][ny] - 'a')),
                    nx * m + ny);
      long long tmp = v.sta * 10000 + v.pos;
      if (s.find(tmp) != s.end()) continue;
      if (visited[nx][ny] && (visited[nx][ny] & v.sta) == visited[nx][ny])
        continue;
      visited[nx][ny] = v.sta;
      s.insert(tmp);
      q.push(v);
    }
  }
  return "-1";
}
int main() {
  string Ans;
  scanf("%d %d %d", &n, &m, &K);
  InputMat();
  Ans = Bfs();
  printf("%s\n", Ans.c_str());
  return 0;
}