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 n, m, k;
char mp[55][55], str[55 * 55];
int prei[55][55], prej[55][55], dis[55][55], vst[55][55];
int q2[2][55 * 55], r2[2], q[55 * 55];
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
bool used[55];
int res = 0x3f3f3f3f, si, ti, sj, tj;
void solve() {
  memset(dis, 0x3f3f3f3f, sizeof(dis));
  dis[ti][tj] = 0;
  int l = 0, r = 0;
  q[r++] = ti * m + tj;
  while (l != r) {
    int mask = q[l++];
    int x = mask / m, y = mask % m;
    for (int i = 0; i < 4; ++i) {
      int nx = x + dx[i], ny = y + dy[i];
      if (nx >= 0 && nx < n && ny >= 0 && ny < m &&
          (mp[nx][ny] == 'S' || mp[nx][ny] == 'T' || used[mp[nx][ny] - 'a']) &&
          dis[x][y] + 1 < dis[nx][ny]) {
        dis[nx][ny] = dis[x][y] + 1;
        q[r++] = nx * m + ny;
      }
    }
  }
  if (dis[si][sj] == 0x3f3f3f3f || dis[si][sj] > res) return;
  bool ok = false;
  if (dis[si][sj] < res) {
    ok = true;
    res = dis[si][sj];
  }
  memset(vst, 0, sizeof(vst));
  bool id = false;
  r2[id] = 0;
  q2[id][r2[id]++] = si * m + sj;
  for (int i = 0; i < dis[si][sj]; ++i) {
    id = !id;
    r2[id] = 0;
    char mini = 'z' + 1;
    for (int j = 0; j < r2[!id]; ++j) {
      int mask = q2[!id][j];
      int x = mask / m, y = mask % m;
      if (mp[x][y] < mini) mini = mp[x][y];
    }
    if (!ok && mini > str[i + 1]) return;
    if (mini < str[i + 1]) ok = true;
    str[i + 1] = mini;
    for (int j = 0; j < r2[!id]; ++j) {
      int mask = q2[!id][j];
      int x = mask / m, y = mask % m;
      if (mp[x][y] == mini) {
        for (int h = 0; h < 4; ++h) {
          int nx = x + dx[h], ny = y + dy[h];
          if (nx >= 0 && nx < n && ny >= 0 && ny < m &&
              dis[x][y] - 1 == dis[nx][ny]) {
            if (vst[nx][ny] != i + 1) {
              vst[nx][ny] = i + 1;
              q2[id][r2[id]++] = nx * m + ny;
            }
          }
        }
      }
    }
  }
}
void dfs(int dep, int count) {
  if (dep == 26) {
    if (count == k) solve();
    return;
  }
  dfs(dep + 1, count);
  if (count < k) {
    used[dep] = true;
    dfs(dep + 1, count + 1);
    used[dep] = false;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%s", mp[i]);
    for (int j = 0; j < m; ++j)
      if (mp[i][j] == 'S')
        si = i, sj = j;
      else if (mp[i][j] == 'T')
        ti = i, tj = j;
  }
  dfs(0, 0);
  if (res == 0x3f3f3f3f)
    puts("-1");
  else {
    for (int i = 2; i <= res; ++i) putchar(str[i]);
    puts("");
  }
  return 0;
}
