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 dy[] = {-1, 0, 1, 0}, dx[] = {0, -1, 0, 1};
inline int D(int x1, int y1, int x2, int y2) {
  return abs(x1 - x2) + abs(y1 - y2);
}
int h, w, k;
string in[50];
void run() {
  cin >> h >> w >> k;
  int sy, sx, gy, gx;
  for (int i = 0; i < h; i++) {
    cin >> in[i];
    for (int j = 0; j < w; j++) {
      if (in[i][j] == 'S')
        sy = i, sx = j;
      else if (in[i][j] == 'T')
        gy = i, gx = j;
    }
  }
  set<pair<int, int> > s;
  priority_queue<pair<int, pair<pair<string, int>, int> >,
                 vector<pair<int, pair<pair<string, int>, int> > >,
                 greater<pair<int, pair<pair<string, int>, int> > > >
      q;
  q.push(
      make_pair(D(sy, sx, gy, gx), make_pair(make_pair("", 0), sy * w + sx)));
  while (!q.empty()) {
    int y = q.top().second.second / w, x = q.top().second.second % w;
    string str = q.top().second.first.first;
    int mask = q.top().second.first.second;
    q.pop();
    if (s.count(make_pair(mask, y * w + x))) continue;
    s.insert(make_pair(mask, y * w + x));
    for (int d = 0; d < 4; d++) {
      int ny = y + dy[d], nx = x + dx[d];
      if (ny < 0 || nx < 0 || ny >= h || nx >= w || ny == sy && nx == sx)
        continue;
      if (ny == gy && nx == gx) {
        cout << str << endl;
        return;
      }
      string next = str + in[ny][nx];
      int nmask = mask | 1 << in[ny][nx] - 'a';
      if (__builtin_popcount(nmask) <= k)
        q.push(make_pair(D(ny, nx, gy, gx) + next.size(),
                         make_pair(make_pair(next, nmask), ny * w + nx)));
    }
  }
  cout << -1 << endl;
}
int main() {
  run();
  return 0;
}
