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 oo = 0x3f3f3f3f;
const double eps = 1e-9;
int R, C, K, SX, SY, TX, TY;
string grid[60], best;
int dist[60][60];
vector<int> dir[60][60];
bool can[26];
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
bool mark[60][60];
void bfs() {
  memset(dist, oo, sizeof(dist));
  for (int i = (0); i < (R); i++)
    for (int j = (0); j < (C); j++) dir[i][j].clear();
  dist[TX][TY] = 0;
  queue<pair<int, int> > q;
  q.push(pair<int, int>(TX, TY));
  while (!q.empty()) {
    int r = q.front().first, c = q.front().second;
    q.pop();
    if (grid[r][c] == 'S') break;
    for (int i = (0); i < (4); i++) {
      int rr = r + dx[i], cc = c + dy[i];
      if (rr < 0 || rr == R || cc < 0 || cc == C) continue;
      if (grid[rr][cc] != 'S' && grid[rr][cc] != 'T' &&
          !can[grid[rr][cc] - 'a'])
        continue;
      if (dist[rr][cc] < dist[r][c] + 1) continue;
      if (dist[rr][cc] > dist[r][c] + 1) {
        dist[rr][cc] = dist[r][c] + 1;
        dir[rr][cc].clear();
        dir[rr][cc].push_back((i + 2) % 4);
        q.push(pair<int, int>(rr, cc));
      } else {
        char ch = grid[rr + dx[dir[rr][cc][0]]][cc + dy[dir[rr][cc][0]]];
        if (ch < grid[r][c]) continue;
        if (ch > grid[r][c]) dir[rr][cc].clear();
        dir[rr][cc].push_back((i + 2) % 4);
      }
    }
  }
  if (dist[SX][SY] == oo) return;
  if (best != "0" && int((best).size()) < dist[SX][SY] - 1) return;
  string cur(dist[SX][SY] - 1, 'z');
  vector<pair<int, int> > level;
  level.push_back(pair<int, int>(SX, SY));
  memset(mark, 0, sizeof(mark));
  mark[SX][SY] = true;
  for (int i = (0); i < (dist[SX][SY] - 1); i++) {
    for (int j = (0); j < (int((level).size())); j++) {
      int r = level[j].first, c = level[j].second;
      cur[i] = min(cur[i], grid[r + dx[dir[r][c][0]]][c + dy[dir[r][c][0]]]);
    }
    vector<pair<int, int> > nxt;
    for (int j = (0); j < (int((level).size())); j++) {
      int r = level[j].first, c = level[j].second;
      if (cur[i] != grid[r + dx[dir[r][c][0]]][c + dy[dir[r][c][0]]]) continue;
      for (int k = (0); k < (int((dir[r][c]).size())); k++) {
        int rr = r + dx[dir[r][c][k]], cc = c + dy[dir[r][c][k]];
        if (!mark[rr][cc]) {
          mark[rr][cc] = true;
          nxt.push_back(pair<int, int>(rr, cc));
        }
      }
    }
    level = nxt;
  }
  if (best == "0" || int((cur).size()) < int((best).size()) ||
      (int((cur).size()) == int((best).size()) && cur < best))
    best = cur;
}
void rek(int p, int k) {
  if (k < 0) return;
  if (p == 26) {
    if (k) return;
    bfs();
  } else {
    can[p] = false;
    rek(p + 1, k);
    can[p] = true;
    rek(p + 1, k - 1);
  }
}
int main() {
  cin >> R >> C >> K;
  for (int i = (0); i < (R); i++) cin >> grid[i];
  for (int i = (0); i < (R); i++)
    for (int j = (0); j < (C); j++) {
      if (grid[i][j] == 'S') {
        SX = i;
        SY = j;
      }
      if (grid[i][j] == 'T') {
        TX = i;
        TY = j;
      }
    }
  best = "0";
  rek(0, K);
  if (best == "0") {
    cout << -1 << endl;
  } else {
    cout << best << endl;
  }
  return 0;
}
