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;
set<int> was;
pair<int, int> pr[55][55];
int ds[55][55];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
char a[55][55];
int n, m, k, si, sj, fi, fj;
string ans = "";
bool ex = 0;
inline bool ok(int i, int j) { return (i >= 0 && i < n && j >= 0 && j < m); }
void go(int mask, int cnt) {
  if (was.find(mask) != was.end()) return;
  was.insert(mask);
  for (int(move) = 0; (move) < (1); ++(move)) {
    for (int(i) = 0; (i) < (n); ++(i))
      for (int(j) = 0; (j) < (m); ++(j)) ds[i][j] = 1000000000;
    queue<pair<int, int> > q;
    q.push(make_pair(fi, fj));
    ds[fi][fj] = 0;
    while (!q.empty()) {
      pair<int, int> cur = q.front();
      q.pop();
      for (int(l) = 0; (l) < (4); ++(l))
        if (ok(cur.first + dx[l], cur.second + dy[l]) &&
            ds[cur.first + dx[l]][cur.second + dy[l]] == 1000000000)
          if (a[cur.first + dx[l]][cur.second + dy[l]] == 'S' ||
              (mask &
               (1 << (a[cur.first + dx[l]][cur.second + dy[l]] - 'a')))) {
            ds[cur.first + dx[l]][cur.second + dy[l]] =
                ds[cur.first][cur.second] + 1;
            q.push(make_pair(cur.first + dx[l], cur.second + dy[l]));
          }
    }
    if (ds[si][sj] == 1000000000) break;
    if (ex && ds[si][sj] - 1 > (int)ans.size()) break;
    string cur = "";
    int step = ds[si][sj];
    for (int(i) = 0; (i) < (n); ++(i))
      for (int(j) = 0; (j) < (m); ++(j))
        if (ds[i][j] == step && (i != si || j != sj)) ds[i][j] = -1;
    while (step > 1) {
      char bst = 'z';
      for (int(i) = 0; (i) < (n); ++(i))
        for (int(j) = 0; (j) < (m); ++(j))
          if (ds[i][j] == step)
            for (int(l) = 0; (l) < (4); ++(l))
              if (ok(i + dx[l], j + dy[l]) &&
                  ds[i + dx[l]][j + dy[l]] == ds[i][j] - 1)
                bst = min(bst, a[i + dx[l]][j + dy[l]]);
      for (int(i) = 0; (i) < (n); ++(i))
        for (int(j) = 0; (j) < (m); ++(j))
          if (ds[i][j] == step - 1) {
            bool can = 0;
            for (int(l) = 0; (l) < (4); ++(l))
              if (ok(i + dx[l], j + dy[l]) && ds[i + dx[l]][j + dy[l]] == step)
                can = 1;
            if (a[i][j] != bst) can = 0;
            if (!can) ds[i][j] = -1;
          }
      cur += bst;
      --step;
    }
    if (!ex || ((int)cur.size() < (int)ans.size()) ||
        ((int)cur.size() == (int)ans.size() && cur < ans)) {
      ex = 1;
      ans = cur;
    }
  }
  if (cnt < k)
    for (int(i) = 0; (i) < (26); ++(i))
      if (!(mask & (1 << i))) go(mask | (1 << i), cnt + 1);
}
int main() {
  cin >> n >> m >> k;
  for (int(i) = 0; (i) < (n); ++(i))
    for (int(j) = 0; (j) < (m); ++(j)) {
      cin >> a[i][j];
      if (a[i][j] == 'S')
        si = i, sj = j;
      else if (a[i][j] == 'T')
        fi = i, fj = j;
    }
  go(0, 0);
  if (ex)
    cout << ans << endl;
  else
    cout << "-1\n";
}
