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;
char str[52][52];
int road[52][52];
int n, m, k;
int sx, sy, ex, ey;
int able[52 * 52];
string bst;
int sequence[52 * 52 * 8];
int afk[52][52];
int vis[52][52];
pair<int, int> pre[52][52];
int ax[] = {-1, 1, 0, 0}, ay[] = {0, 0, -1, 1};
void trysolve() {
  memset(vis, 0x3f, sizeof vis);
  memset(afk, 0x3f, sizeof afk);
  queue<pair<int, int> > q[2];
  int now = 0;
  vis[sx][sy] = 1;
  q[now].push(make_pair(sx, sy));
  afk[sx][sy] = 1;
  while (q[now].size() and vis[ex][ey] > 1e8) {
    vector<pair<int, int> > all;
    while (q[now].size()) {
      int x = q[now].front().first, y = q[now].front().second;
      q[now].pop();
      for (int i = 0; i < 4; i++) {
        int tx = x + ax[i], ty = y + ay[i];
        if (str[tx][ty] != 'T' and !able[road[tx][ty]]) continue;
        int ene = afk[x][y] * 5 + able[road[tx][ty]];
        if (vis[tx][ty] > vis[x][y] + 1) {
          vis[tx][ty] = vis[x][y] + 1;
          afk[tx][ty] = ene;
          all.push_back(make_pair(tx, ty));
          pre[tx][ty] = make_pair(x, y);
        } else if (vis[tx][ty] == vis[x][y] + 1 and afk[tx][ty] > ene) {
          afk[tx][ty] = ene;
          pre[tx][ty] = make_pair(x, y);
        }
      }
    }
    int upp = 0;
    for (int i = 0; i < all.size(); i++)
      sequence[afk[all[i].first][all[i].second]] = 1,
      upp = max(upp, afk[all[i].first][all[i].second]);
    int tot = 0;
    for (int i = 1; i <= upp; i++)
      if (sequence[i]) sequence[i] = ++tot;
    for (int i = 0; i < all.size(); i++)
      afk[all[i].first][all[i].second] =
          sequence[afk[all[i].first][all[i].second]];
    for (int i = 1; i <= upp; i++)
      if (sequence[i]) sequence[i] = 0;
    now ^= 1;
    for (int i = 0; i < all.size(); i++) q[now].push(all[i]);
  }
  if (vis[ex][ey] < 1e8) {
    int x = pre[ex][ey].first, y = pre[ex][ey].second;
    string gg;
    while (x != sx or y != sy) {
      gg += char('a' - 1 + road[x][y]);
      pair<int, int> op = pre[x][y];
      x = op.first, y = op.second;
    }
    reverse(gg.begin(), gg.end());
    if (bst.size() > gg.size()) bst = gg;
    if (bst.size() == gg.size()) bst = min(bst, gg);
  }
}
void solve(int xk, int cnt) {
  if (cnt > k) return;
  if (xk == 27) trysolve();
  if (xk == 27) return;
  solve(xk + 1, cnt);
  able[xk] = cnt + 1;
  solve(xk + 1, cnt + 1);
  able[xk] = 0;
}
signed main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%s", str[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (str[i][j] == 'S') sx = i, sy = j;
      if (str[i][j] == 'T') ex = i, ey = j;
      bst += 'o';
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) road[i][j] = str[i][j] - 'a' + 1;
  solve(1, 0);
  if (bst.size() >= n * m) bst = "-1";
  cout << bst;
  return 0;
}
