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 INF = 0x3f3f3f3f;
char field[55][55];
int n, m, k;
int can;
int sx, sy, tx, ty;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
string ans;
vector<char> vt;
bool out(int x, int y) { return x < 0 || x >= n || y < 0 || y >= m; }
void cal() {
  int dist[55][55];
  fill(dist[0], dist[n], INF);
  queue<pair<int, int> > Q;
  Q.push(make_pair(tx, ty));
  dist[tx][ty] = 0;
  while (!Q.empty()) {
    pair<int, int> now = Q.front(), next;
    Q.pop();
    if (now == make_pair(sx, sy)) {
      break;
    }
    for (int i = 0; i < 4; i++) {
      next = now;
      next.first += dx[i];
      next.second += dy[i];
      if (out(next.first, next.second) ||
          dist[next.first][next.second] != INF) {
        continue;
      }
      if (next != make_pair(sx, sy) &&
          (can & (1 << field[next.first][next.second] - 'a')) == 0) {
        continue;
      }
      dist[next.first][next.second] = dist[now.first][now.second] + 1;
      Q.push(next);
    }
  }
  if (dist[sx][sy] == INF) {
    return;
  }
  int len = dist[sx][sy] - 1;
  if (ans != "-1" && len > ans.length()) {
    return;
  }
  string path;
  vector<pair<int, int> > vt[2];
  int p = 0, q = 1;
  vt[p].push_back(make_pair(sx, sy));
  for (int i = 0; i < len; i++) {
    vt[q].clear();
    for (vector<pair<int, int> >::iterator it = vt[p].begin();
         it != vt[p].end(); it++) {
      pair<int, int> now = *it, next;
      for (int i = 0; i < 4; i++) {
        next = now;
        next.first += dx[i];
        next.second += dy[i];
        if (out(next.first, next.second) ||
            dist[next.first][next.second] != dist[now.first][now.second] - 1) {
          continue;
        }
        if (vt[q].empty()) {
          vt[q].push_back(next);
        } else {
          pair<int, int> &f = vt[q].front();
          if (field[f.first][f.second] > field[next.first][next.second]) {
            vt[q].clear();
            vt[q].push_back(next);
          } else if (field[f.first][f.second] ==
                     field[next.first][next.second]) {
            vt[q].push_back(next);
          }
        }
      }
    }
    sort(vt[q].begin(), vt[q].end());
    vt[q].resize(unique(vt[q].begin(), vt[q].end()) - vt[q].begin());
    pair<int, int> &f = vt[q].front();
    path += field[f.first][f.second];
    swap(p, q);
  }
  if (ans == "-1" || len < ans.length()) {
    ans = path;
    return;
  }
  if (ans > path) {
    ans = path;
  }
}
void dfs(int pos, int have) {
  if (have == k) {
    cal();
    return;
  }
  for (int i = pos; i < vt.size(); i++) {
    can ^= (1 << vt[i]);
    dfs(i + 1, have + 1);
    can ^= (1 << vt[i]);
  }
}
int main() {
  for (; ~scanf("%d%d%d", &n, &m, &k);) {
    vt.clear();
    for (int i = 0; i < n; i++) {
      scanf("%s", field[i]);
      for (int j = 0; j < m; j++) {
        if (field[i][j] == 'S') {
          sx = i;
          sy = j;
        } else if (field[i][j] == 'T') {
          tx = i;
          ty = j;
        } else {
          vt.push_back(field[i][j] - 'a');
        }
      }
    }
    sort(vt.begin(), vt.end());
    vt.resize(unique(vt.begin(), vt.end()) - vt.begin());
    k = min(k, (int)vt.size());
    ans = "-1";
    can = 0;
    dfs(0, 0);
    cout << ans << endl;
  }
  return 0;
}
