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 Inf = 1000000000;
const int Maxn = 50;
const int Maxl = 26;
const int Maxd = 4;
const int dy[Maxd] = {1, 0, -1, 0};
const int dx[Maxd] = {0, 1, 0, -1};
int n, m, k;
char B[Maxn][Maxn];
int sx, sy, ex, ey;
int has[Maxl];
int dist[Maxn][Maxn];
pair<int, int> p[Maxn][Maxn];
string s, res;
bool sol;
void BFS() {
  fill((int *)dist, (int *)dist + Maxn * Maxn, Inf);
  pair<int, int> v = make_pair(ex, ey);
  dist[v.first][v.second] = 0;
  vector<pair<int, int> > seq, buck[Maxl];
  seq.push_back(v);
  for (int i = 0; !seq.empty(); i++) {
    for (int j = 0; j < seq.size(); j++) {
      v = seq[j];
      for (int d = 0; d < Maxd; d++) {
        int ni = v.first + dy[d], nj = v.second + dx[d];
        if (0 <= ni && ni < n && 0 <= nj && nj < m && i + 1 < dist[ni][nj] &&
            (B[ni][nj] == 'S' || has[B[ni][nj] - 'a'])) {
          dist[ni][nj] = i + 1;
          p[ni][nj] = make_pair(v.first, v.second);
          if (B[ni][nj] == 'S')
            return;
          else
            buck[B[ni][nj] - 'a'].push_back(make_pair(ni, nj));
        }
      }
    }
    seq.clear();
    for (int j = 0; j < Maxl; j++) {
      for (int l = 0; l < buck[j].size(); l++) seq.push_back(buck[j][l]);
      buck[j].clear();
    }
  }
}
void getPath(string &s) {
  s = "";
  pair<int, int> v = p[sx][sy];
  while (v.first != ex || v.second != ey) {
    s += B[v.first][v.second];
    v = p[v.first][v.second];
  }
}
void Search(int v, int from) {
  if (v == k) {
    BFS();
    if (dist[sx][sy] == Inf) return;
    getPath(s);
    if (!sol || s.length() < res.length() ||
        s.length() == res.length() && s < res)
      res = s;
    sol = true;
  } else
    for (int i = from; i < Maxl; i++) {
      has[i] = true;
      Search(v + 1, i + 1);
      has[i] = false;
    }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      scanf(" %c", &B[i][j]);
      if (B[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (B[i][j] == 'T') {
        ex = i;
        ey = j;
        B[i][j] = 'a';
      }
    }
  Search(0, 0);
  if (!sol)
    printf("-1\n");
  else
    printf("%s\n", res.c_str());
  return 0;
}
