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>
#pragma comment(linker, "/STACK:60000000")
using namespace std;
const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
template <class T>
T sqr(T x) {
  return x * x;
}
int n, m, k;
char a[50][50];
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
bool u[256];
int d[50][50];
pair<int, int> q[2500];
string bfs() {
  memset(d, 255, sizeof(d));
  pair<int, int> S, T;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++) {
      if (a[i][j] == 'S') S = make_pair(i, j);
      if (a[i][j] == 'T') T = make_pair(i, j);
    }
  d[T.first][T.second] = 0;
  int h, t;
  h = t = 0;
  q[h] = T;
  while (h <= t) {
    pair<int, int> v = q[h];
    h++;
    for (int i = 0; i < (int)(4); i++) {
      pair<int, int> w = make_pair(v.first + dx[i], v.second + dy[i]);
      if (0 <= w.first && w.first < n && 0 <= w.second && w.second < m &&
          d[w.first][w.second] == -1 && u[a[w.first][w.second]]) {
        d[w.first][w.second] = d[v.first][v.second] + 1;
        t++;
        q[t] = w;
      }
    }
  }
  if (d[S.first][S.second] == -1) return "S";
  string res = "";
  vector<pair<int, int> > V, W;
  V.push_back(S);
  int cd = d[S.first][S.second];
  while (cd > 1) {
    char c = 'z';
    for (int i = 0; i < (int)(V.size()); i++)
      for (int j = 0; j < (int)(4); j++) {
        pair<int, int> w = make_pair(V[i].first + dx[j], V[i].second + dy[j]);
        if (0 <= w.first && w.first < n && 0 <= w.second && w.second < m &&
            cd - 1 == d[w.first][w.second])
          c = min(c, a[w.first][w.second]);
      }
    res += c;
    W.clear();
    for (int i = 0; i < (int)(V.size()); i++)
      for (int j = 0; j < (int)(4); j++) {
        pair<int, int> w = make_pair(V[i].first + dx[j], V[i].second + dy[j]);
        if (0 <= w.first && w.first < n && 0 <= w.second && w.second < m &&
            cd - 1 == d[w.first][w.second] && a[w.first][w.second] == c)
          W.push_back(w);
      }
    sort(W.begin(), W.end());
    W.resize(unique(W.begin(), W.end()) - W.begin());
    cd--;
    V = W;
    W.clear();
  }
  return res;
}
string res;
void calc(int cur, int la) {
  if (cur == 26) {
    string tmp = bfs();
    if (tmp == "S") return;
    if (res == "S")
      res = tmp;
    else if (res.length() > tmp.length() ||
             (res.length() == tmp.length() && res > tmp))
      res = tmp;
    return;
  }
  calc(cur + 1, la);
  if (la == 0) return;
  u[cur + 'a'] = 1;
  calc(cur + 1, la - 1);
  u[cur + 'a'] = 0;
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++) cin >> a[i][j];
  memset(u, 0, sizeof(u));
  u['S'] = 1;
  u['T'] = 1;
  res = "S";
  calc(0, k);
  if (res == "S")
    cout << -1 << endl;
  else
    cout << res << endl;
  return 0;
}
