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;
char in[55][55];
int used[300];
char ap[100], apc;
string ans;
int sx, sy;
int ex, ey;
int all[300], ac;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int que[3000][2];
int vis[55][55];
int dis[55][55];
int visid;
vector<pair<int, int> > pos[2000];
inline void solve() {
  int i, j;
  visid++;
  int qs = 0, qe = 0;
  que[qe][0] = sx;
  que[qe][1] = sy;
  qe++;
  vis[sx][sy] = visid;
  dis[sx][sy] = 0;
  while (qs < qe) {
    int x = que[qs][0], y = que[qs][1];
    qs++;
    for (i = 0; i < ac; i++) {
      for (j = 0; j < 4; j++) {
        int xx = x + dx[j], yy = y + dy[j];
        if (xx >= 0 && xx < n && yy >= 0 && yy < m && vis[xx][yy] != visid &&
            in[xx][yy] == all[i]) {
          que[qe][0] = xx;
          que[qe][1] = yy;
          qe++;
          vis[xx][yy] = visid;
          dis[xx][yy] = dis[x][y] + 1;
        }
      }
    }
    for (j = 0; j < 4; j++) {
      int xx = x + dx[j], yy = y + dy[j];
      if (xx >= 0 && xx < n && yy >= 0 && yy < m && vis[xx][yy] != visid &&
          in[xx][yy] == 'S') {
        que[qe][0] = xx;
        que[qe][1] = yy;
        qe++;
        vis[xx][yy] = visid;
        dis[xx][yy] = dis[x][y] + 1;
      }
    }
  }
  if (vis[ex][ey] != visid) return;
  string str;
  int d = dis[ex][ey];
  pos[d].clear();
  pos[d].push_back(make_pair(ex, ey));
  visid++;
  vis[ex][ey] = visid;
  while (d > 1) {
    char c = 'z' + 1;
    for (i = 0; i < pos[d].size(); i++) {
      int x = pos[d][i].first, y = pos[d][i].second;
      for (j = 0; j < 4; j++) {
        int xx = x + dx[j], yy = y + dy[j];
        if (xx >= 0 && xx < n && yy >= 0 && yy < m &&
            vis[xx][yy] == visid - 1 && dis[xx][yy] == dis[x][y] - 1) {
          vis[xx][yy] = visid;
          if (in[xx][yy] < c) {
            c = in[xx][yy];
            pos[d - 1].clear();
          }
          if (in[xx][yy] == c) {
            pos[d - 1].push_back(make_pair(xx, yy));
          }
        }
      }
    }
    str += c;
    d--;
  }
  if (str.size() < ans.size() || (str.size() == ans.size() && str < ans))
    ans = str;
}
inline void dfs(int np, int c) {
  solve();
  if (np == apc) return;
  dfs(np + 1, c);
  if (c < k) {
    all[ac++] = ap[np];
    dfs(np + 1, c + 1);
    ac--;
  }
}
int main() {
  int i, j;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i < n; i++) scanf("%s", in[i]);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (in[i][j] == 'S') ex = i, ey = j;
      if (in[i][j] == 'T') sx = i, sy = j;
      used[in[i][j]] = 1;
    }
  for (i = 'a'; i <= 'z'; i++)
    if (used[i]) ap[apc++] = i;
  for (int i = 0; i < 1000; i++) ans += 'z';
  dfs(0, 0);
  if (ans.size() == 1000)
    puts("-1");
  else
    puts(ans.c_str());
}
