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 N = 51, A = 26;
int n, m, k, dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1}, mark[A];
vector<pair<int, int> > q1, q2;
pair<int, int> s, t;
string dis[N][N], a[N], ans;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (~a[i].find('S')) s = {i, a[i].find('S')};
    if (~a[i].find('T')) t = {i, a[i].find('T')};
  }
  fill(mark, mark + k, 1);
  sort(mark, mark + A);
  do {
    for (q1.push_back(s), dis[s.first][s.second] = "S"; q1.size(); q2.clear()) {
      sort(q1.begin(), q1.end(), [](pair<int, int> x, pair<int, int> y) {
        return dis[x.first][x.second] < dis[y.first][y.second];
      });
      for (auto [x, y] : q1)
        for (int i = 0; i < 4; i++) {
          int z = x + dx[i], t = y + dy[i];
          if (~z && z < n && ~t && t < m && dis[z][t] == "" &&
              (a[z][t] == 'T' || mark[a[z][t] - 'a'])) {
            q2.push_back({z, t});
            dis[z][t] = dis[x][y] + a[z][t];
          }
        }
      q1 = q2;
    }
    string tmp = dis[t.first][t.second];
    if (tmp != "" && (ans == "" || ans.size() > tmp.size() ||
                      (ans.size() == tmp.size() && ans > tmp)))
      ans = tmp;
    fill(*dis, *end(dis), "");
  } while (next_permutation(mark, mark + A));
  cout << (ans == "" ? "-1" : ans.substr(1, ans.size() - 2)) << endl;
}
