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;
int k;
char str[55][55];
int sr, sc;
int er, ec;
string cur;
bool can[26];
int am = 0;
int mark[55][55];
int dist[55][55];
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int qhead, qtail;
int q[55 * 55 * 4];
int nqtail;
int nq[55 * 55 * 4];
int go(int r, int c) {
  qhead = qtail = 0;
  mark[r][c] = am;
  dist[r][c] = 0;
  q[qtail++] = r;
  q[qtail++] = c;
  while (qhead < qtail) {
    int cr = q[qhead++];
    int cc = q[qhead++];
    if (str[cr][cc] == 'S') return dist[cr][cc];
    for (int dir = 0; dir < 4; ++dir) {
      int nr = cr + dx[dir];
      int nc = cc + dy[dir];
      if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
      if (mark[nr][nc] == am) continue;
      if (str[nr][nc] != 'S' && (str[nr][nc] == 'T' || !can[str[nr][nc] - 'a']))
        continue;
      mark[nr][nc] = am;
      dist[nr][nc] = dist[cr][cc] + 1;
      q[qtail++] = nr;
      q[qtail++] = nc;
    }
  }
  return 987654321;
}
void rec(int r, int c) {
  qhead = qtail = 0;
  nqtail = 0;
  q[qtail++] = r;
  q[qtail++] = c;
  mark[r][c] = am;
  cur = "";
  for (int stp = 0; stp < dist[r][c] - 1; ++stp) {
    int best = 255;
    for (int i = 0; i < qtail; i += 2) {
      int cr = q[i];
      int cc = q[i + 1];
      for (int dir = 0; dir < 4; ++dir) {
        int nr = cr + dx[dir];
        int nc = cc + dy[dir];
        if (mark[nr][nc] == am) continue;
        if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
        if (dist[nr][nc] != dist[cr][cc] - 1) continue;
        if (str[nr][nc] != 'T' &&
            (str[nr][nc] == 'S' || !can[str[nr][nc] - 'a']))
          continue;
        best = min(best, (int)str[nr][nc]);
      }
    }
    cur += char(best);
    for (int i = 0; i < qtail; i += 2) {
      int cr = q[i];
      int cc = q[i + 1];
      for (int dir = 0; dir < 4; ++dir) {
        int nr = cr + dx[dir];
        int nc = cc + dy[dir];
        if (mark[nr][nc] == am) continue;
        if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
        if (dist[nr][nc] != dist[cr][cc] - 1) continue;
        if (str[nr][nc] != 'T' &&
            (str[nr][nc] == 'S' || !can[str[nr][nc] - 'a']))
          continue;
        if (best == str[nr][nc]) {
          mark[nr][nc] = am;
          nq[nqtail++] = nr;
          nq[nqtail++] = nc;
        }
      }
    }
    for (int i = 0; i < nqtail; ++i) {
      q[i] = nq[i];
    }
    qtail = nqtail;
    nqtail = 0;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    cin >> str[i];
    for (int j = 0; j < m; ++j)
      if (str[i][j] == 'S') {
        sr = i;
        sc = j;
      } else if (str[i][j] == 'T') {
        er = i;
        ec = j;
      }
  }
  int dist = 987654321;
  string res = "";
  memset(can, 0, sizeof(can));
  memset(mark, -1, sizeof(mark));
  for (int a = 0; a < 26; ++a) {
    for (int b = a; b < 26; ++b) {
      for (int c = b; c < 26; ++c) {
        for (int d = c; d < 26; ++d) {
          int cnt = 1;
          if (b != a) ++cnt;
          if (c != b) ++cnt;
          if (d != c) ++cnt;
          if (cnt > k) continue;
          ++am;
          cur = "";
          can[a] = 1;
          can[b] = 1;
          can[c] = 1;
          can[d] = 1;
          int ndist = go(er, ec);
          if (ndist <= dist && ndist != 987654321) {
            ++am;
            rec(sr, sc);
            if (ndist < dist || ndist == dist && cur < res) {
              dist = ndist;
              res = cur;
            }
          }
          can[a] = 0;
          can[b] = 0;
          can[c] = 0;
          can[d] = 0;
        }
      }
    }
  }
  if (dist == 987654321)
    cout << "-1" << endl;
  else
    cout << res << endl;
  return 0;
}
