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 dx[4] = {1, 0, -1, 0};
const int dy[5] = {0, 1, 0, -1};
string res = "";
int que[55 * 55][2], _que[55 * 55][2];
int dist[55][55];
bool p[55][55];
char g[55][55];
int n, m, limit;
int sx, sy, tx, ty;
bool flag[100];
int main() {
  scanf("%d%d%d", &n, &m, &limit);
  for (int i = 0; i < n; i++) scanf("%s", g[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (g[i][j] == 'S')
        sx = i, sy = j;
      else if (g[i][j] == 'T')
        tx = i, ty = j;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (g[i][j] >= 'a' && g[i][j] <= 'z') flag[g[i][j] - 'a'] = 1;
  flag[26] = 1;
  if (abs(sx - tx) + abs(sy - ty) == 1) return 0;
  for (int i = 0; i < 26; i++)
    if (flag[i])
      for (int j = i + 1; j < 27; j++)
        if (flag[j]) {
          if (j < 26 && limit == 1) continue;
          for (int k = j; k < 27; k++)
            if (flag[k]) {
              if (k < 26 && limit == 2) continue;
              if (k == j && k < 26) continue;
              for (int l = k; l < 27; l++)
                if (flag[l]) {
                  if (l < 26 && limit == 3) continue;
                  if (l == k && l < 26) continue;
                  int XXX = (1 << i) + (1 << j) + (1 << k) + (1 << l);
                  for (int a = 0; a < n; a++)
                    for (int b = 0; b < m; b++) dist[a][b] = -1, p[a][b] = 0;
                  dist[tx][ty] = 0;
                  que[0][0] = tx;
                  que[0][1] = ty;
                  int r = 0;
                  for (int l = 0; dist[sx][sy] == -1 && l <= r; l++) {
                    int u = que[l][0], v = que[l][1];
                    for (int w = 0; w < 4; w++)
                      if (u + dx[w] >= 0 && u + dx[w] < n && v + dy[w] >= 0 &&
                          v + dy[w] < m && dist[u + dx[w]][v + dy[w]] == -1) {
                        if (sx == u + dx[w] && sy == v + dy[w]) {
                          dist[sx][sy] = dist[u][v] + 1;
                          continue;
                        }
                        int uu = u + dx[w], vv = v + dy[w];
                        if (XXX & (1 << (g[uu][vv] - 'a'))) {
                          dist[uu][vv] = dist[u][v] + 1;
                          que[++r][0] = uu;
                          que[r][1] = vv;
                        }
                      }
                  }
                  if (dist[sx][sy] != -1) {
                    string tmp = "";
                    int r = 0;
                    que[0][0] = sx;
                    que[0][1] = sy;
                    while (tmp.size() != dist[sx][sy] - 1) {
                      char ww = 'z' + 1;
                      for (int l = 0; l <= r; l++) {
                        int u = que[l][0], v = que[l][1];
                        for (int w = 0; w < 4; w++) {
                          int uu = u + dx[w], vv = v + dy[w];
                          if (uu >= 0 && uu < n && vv >= 0 && vv < m &&
                              dist[uu][vv] != -1 &&
                              dist[uu][vv] + tmp.size() == dist[sx][sy] - 1)
                            ww = min(ww, g[uu][vv]);
                        }
                      }
                      int _r = -1;
                      for (int l = 0; l <= r; l++) {
                        int u = que[l][0], v = que[l][1];
                        for (int w = 0; w < 4; w++) {
                          int uu = u + dx[w], vv = v + dy[w];
                          if (uu >= 0 && uu < n && vv >= 0 && vv < m &&
                              !p[uu][vv] && dist[uu][vv] != -1 &&
                              dist[uu][vv] + tmp.size() == dist[sx][sy] - 1 &&
                              g[uu][vv] == ww) {
                            p[uu][vv] = 1;
                            _que[++_r][0] = uu;
                            _que[_r][1] = vv;
                          }
                        }
                      }
                      r = _r;
                      for (int l = 0; l <= r; l++)
                        que[l][0] = _que[l][0], que[l][1] = _que[l][1];
                      tmp += ww;
                    }
                    if (!res.size() || (res.size() > tmp.size()) ||
                        (res.size() == tmp.size() && res > tmp))
                      res = tmp;
                  }
                }
            }
        }
  if (!res.size())
    printf("-1\n");
  else
    cout << res << endl;
}
