Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
int main() {
  long long n, m, k, sx, sy, tx, ty;
  cin >> n >> m >> k;
  vector<string> mp(n);
  for (long long i = 0; i < ((long long)n); i++) cin >> mp[i];
  for (long long i = 0; i < ((long long)n); i++)
    for (long long j = 0; j < ((long long)m); j++)
      if (mp[i][j] == 'S') sx = i, sy = j;
  for (long long i = 0; i < ((long long)n); i++)
    for (long long j = 0; j < ((long long)m); j++)
      if (mp[i][j] == 'T') tx = i, ty = j;
  priority_queue<
      pair<int, pair<pair<string, int>, pair<int, int> > >,
      vector<pair<int, pair<pair<string, int>, pair<int, int> > > >,
      greater<pair<int, pair<pair<string, int>, pair<int, int> > > > >
      q;
  q.push(make_pair((abs((sx) - (tx)) + abs((sy) - (ty))),
                   make_pair(make_pair("", 0), make_pair(sx, sy))));
  set<pair<int, pair<int, int> > > s;
  while (((long long)q.size())) {
    pair<int, pair<pair<string, int>, pair<int, int> > > pp = q.top();
    pair<pair<string, int>, pair<int, int> > p = pp.second;
    q.pop();
    string str = p.first.first;
    int mask = p.first.second;
    int x = p.second.first;
    int y = p.second.second;
    if (s.find(make_pair(mask, make_pair(x, y))) != s.end()) continue;
    s.insert(make_pair(mask, make_pair(x, y)));
    for (long long i = 0; i < ((long long)4); i++) {
      int xx = x + dx[i];
      int yy = y + dy[i];
      if (xx < 0 || n <= xx || yy < 0 || m <= yy) continue;
      if (mp[xx][yy] == 'T') {
        cout << str << endl;
        return 0;
      }
      if ('a' <= mp[xx][yy] && mp[xx][yy] <= 'z') {
        int next = (1 << (mp[xx][yy] - 'a')) | mask;
        if (k < __builtin_popcount(next)) continue;
        q.push(make_pair(
            ((long long)str.size()) + 1 + (abs((xx) - (tx)) + abs((yy) - (ty))),
            make_pair(make_pair(str + string(1, mp[xx][yy]), next),
                      make_pair(xx, yy))));
      }
    }
  }
  cout << "-1" << endl;
}