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;
int n, m, K, S, T;
int dis(int a, int b) { return abs(a / m - b / m) + abs(a % m - b % m); }
string s[55];
int main() {
  cin >> n >> m >> K;
  for (int i = 0; i < n; ++i) cin >> s[i];
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (s[i][j] == 'S') S = i * m + j;
      if (s[i][j] == 'T') T = i * m + j;
    }
  priority_queue<pair<pair<int, string>, pair<int, int> >,
                 vector<pair<pair<int, string>, pair<int, int> > >,
                 greater<pair<pair<int, string>, pair<int, int> > > >
      Q;
  set<pair<int, int> > Set;
  Q.push(make_pair(make_pair(dis(S, T), ""), make_pair(0, S)));
  while (!Q.empty()) {
    pair<pair<int, string>, pair<int, int> > c = Q.top();
    Q.pop();
    string ss = c.first.second;
    int mask = c.second.first, id = c.second.second;
    if (Set.count(make_pair(mask, id))) continue;
    Set.insert(make_pair(mask, id));
    int x = id / m, y = id % m;
    for (int d = 0; d < 4; ++d) {
      int nx = x + (d == 2) - (d == 0), ny = y + (d == 3) - (d == 1);
      if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
      if (nx * m + ny == T) {
        cout << ss << endl;
        return 0;
      }
      string nss = ss + s[nx][ny];
      int nmask = mask | (1 << (s[nx][ny] - 'a'));
      if (__builtin_popcount(nmask) <= K)
        Q.push(
            make_pair(make_pair(dis(nx * m + ny, T) + (int)(nss).size(), nss),
                      make_pair(nmask, nx * m + ny)));
    }
  }
  cout << -1 << endl;
  return 0;
}
