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 n, m, k, S, T, dir[4] = {1, -1};
char *grid;
inline int dis(int i, int j) { return abs(i / m - j / m) + abs(i % m - j % m); }
string dijkstra() {
  priority_queue<tuple<int, string, int, int>,
                 vector<tuple<int, string, int, int> >,
                 greater<tuple<int, string, int, int> > >
      q;
  set<pair<int, int> > vis;
  q.push(make_tuple(dis(S, T), "", 0, S));
  while (!q.empty()) {
    auto c = q.top();
    string path = get<1>(c);
    int i = get<3>(c);
    int mask = get<2>(c);
    q.pop();
    if (vis.insert(make_pair(i, mask)).second)
      for (int j = 0; j < 4; j++) {
        int t = mask | (1 << (grid[i + dir[j]] - 'a'));
        if (grid[i + dir[j]] == 'S')
          continue;
        else if (grid[i + dir[j]] == 'T')
          return path;
        else if (__builtin_popcount(t) <= k &&
                 !vis.count(make_pair(i + dir[j], t)))
          q.push(make_tuple(dis(i + dir[j], T) + path.size() + 1,
                            path + grid[i + dir[j]], t, i + dir[j]));
      }
  }
  return "-1";
}
int main() {
  cin >> n >> m >> k;
  n += 2, dir[3] = -(dir[2] = m += 2);
  grid = (char *)malloc(n * m);
  memset(grid, 'S', n * m);
  for (int i = 1; i < n - 1; i++)
    for (int j = 1; j < m - 1; j++) {
      cin >> grid[i * m + j];
      if (grid[i * m + j] == 'S') S = i * m + j;
      if (grid[i * m + j] == 'T') T = i * m + j;
    }
  cout << dijkstra() << endl;
  return 0;
}
