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;
int S, T;
string s[55];
set<pair<int, int>> se;
int dist(int a, int b) { return abs(a / m - b / m) + abs(a % m - b % m); }
int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
void Dijstra() {
  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;
  Q.push({{dist(S, T), ""}, {0, S}});
  pair<pair<int, string>, pair<int, int>> ss;
  string str;
  int book, pos;
  int x, y, xx, yy;
  while (!Q.empty()) {
    ss = Q.top();
    Q.pop();
    str = ss.first.second;
    book = ss.second.first;
    pos = ss.second.second;
    if (!se.count({book, pos})) {
      se.insert({book, pos});
      x = pos / m, y = pos % m;
      for (int i = 0; i < 4; i++) {
        xx = x + dir[i][0];
        yy = y + dir[i][1];
        if (xx >= 0 && xx < n && yy >= 00 && yy < m) {
          if (s[xx][yy] == 'T') {
            cout << str << endl;
            exit(0);
          }
          if (__builtin_popcount(book | (1 << (s[xx][yy] - 'a'))) <= k)
            Q.push({{dist(xx * m + yy, T) + str.length() + 1, str + s[xx][yy]},
                    {book | (1 << (s[xx][yy] - 'a')), xx * m + yy}});
        }
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> s[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;
      }
    }
  }
  Dijstra();
  cout << -1 << endl;
}
