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>
int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
char mm[64][64];
char path[2500];
int d[64][64], pre[64][64];
int n, m, K;
int ans;
int tx, ty, sx, sy;
int st[4];
int MM;
int bs[2][4][2500];
int cnt[2][4];
int mmm[32];
int valid(int x, int y) {
  return x >= 0 && y >= 0 && x < n && y < m &&
         (mm[x][y] < 'a' || mmm[mm[x][y] - 'a'] >= 0) && d[x][y] == 1000000000;
}
void bfs(int mask) {
  int i, j, tail = 0, h, k, l, t, p;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) d[i][j] = 1000000000, pre[i][j] = -1;
  memset(mmm, 0xff, sizeof(mmm));
  for (i = j = 0; i < 32; i++)
    if (mask & (1 << i)) mmm[i] = j++;
  d[tx][ty] = 0;
  for (i = 0; i < 2; i++)
    for (j = 0; j < 4; j++) cnt[i][j] = 0;
  for (i = 0; i < 4; i++)
    if (valid(tx + dir[i][0], ty + dir[i][1])) {
      d[tx + dir[i][0]][ty + dir[i][1]] = 1;
      pre[tx + dir[i][0]][ty + dir[i][1]] = tx << 6 | ty;
      k = mmm[mm[tx + dir[i][0]][ty + dir[i][1]] - 'a'];
      bs[1][k][cnt[1][k]++] = (tx + dir[i][0]) << 6 | (ty + dir[i][1]);
    }
  for (h = 1;
       d[sx][sy] == 1000000000 && cnt[h][0] + cnt[h][1] + cnt[h][2] + cnt[h][3];
       h = !h) {
    cnt[!h][0] = cnt[!h][1] = cnt[!h][2] = cnt[!h][3] = 0;
    for (k = 0; k < 4; k++)
      for (l = 0; l < cnt[h][k]; l++) {
        i = bs[h][k][l] >> 6, j = bs[h][k][l] & 63;
        for (t = 0; t < 4; t++)
          if (valid(i + dir[t][0], j + dir[t][1])) {
            d[i + dir[t][0]][j + dir[t][1]] = d[i][j] + 1;
            pre[i + dir[t][0]][j + dir[t][1]] = i << 6 | j;
            p = mmm[mm[i + dir[t][0]][j + dir[t][1]] - 'a'];
            bs[!h][p][cnt[!h][p]++] = (i + dir[t][0]) << 6 | (j + dir[t][1]);
          }
      }
  }
  if (d[sx][sy] == 1000000000 || d[sx][sy] > ans) return;
  if (ans == d[sx][sy]) {
    for (i = sx, j = sy, k = 0; k < ans; k++) {
      if (path[k] < mm[i][j])
        return;
      else if (path[k] > mm[i][j])
        break;
      h = pre[i][j];
      i = h >> 6, j = h & 63;
    }
  }
  ans = d[sx][sy];
  for (i = sx, j = sy, k = 0; k < ans; k++) {
    path[k] = mm[i][j];
    h = pre[i][j];
    i = h >> 6, j = h & 63;
  }
}
void dfs(int t, int b, int mask) {
  int i;
  if (t == K || mask == MM) {
    bfs(mask);
    return;
  }
  for (i = b; i < 26; i++)
    if (MM & 1 << i) dfs(t + 1, i + 1, mask | 1 << i);
}
int main() {
  int i, j;
  scanf("%d %d %d", &n, &m, &K);
  for (i = 0; i < n; i++) {
    scanf("%s", mm[i]);
    for (j = 0; j < m; j++) {
      if (mm[i][j] == 'S')
        sx = i, sy = j;
      else if (mm[i][j] == 'T')
        tx = i, ty = j;
      else
        MM |= 1 << (mm[i][j] - 'a');
    }
  }
  if (abs(sx - tx) + abs(sy - ty) == 1) {
    puts("");
    return 0;
  }
  ans = 1000000000;
  dfs(0, 0, 0);
  if (ans == 1000000000)
    puts("-1");
  else {
    for (i = 1; i < ans; i++) putchar(path[i]);
    puts("");
  }
  return 0;
}
