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 N = 55;
int n, m, K, sx, sy, ex, ey;
char s[N][N];
vector<char> has;
vector<int> ok;
bool fd = false;
string ans;
int len = 1 << 20;
pair<int, int> pre[N][N];
int rk[N][N];
bool O[300];
const int inf = 1 << 30;
bool Cmp(const pair<int, int>& a, const pair<int, int>& b) {
  if (rk[a.first][a.second] != rk[b.first][b.second])
    return rk[a.first][a.second] < rk[b.first][b.second];
  return s[a.first][a.second] < s[b.first][b.second];
}
bool same(const pair<int, int>& a, const pair<int, int>& b) {
  return s[a.first][a.second] == s[b.first][b.second] &&
         rk[a.first][a.second] == rk[b.first][b.second];
}
int tmp[N * N];
void solve() {
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) rk[i][j] = inf;
  vector<pair<int, int> > Q;
  Q.push_back(make_pair(sx, sy));
  rk[sx][sy] = 0;
  for (int i = 0; i < (int)(ok).size(); ++i) O[ok[i]] = true;
  int L = 0;
  O[(int)'T'] = true;
  while ((int)(Q).size()) {
    vector<pair<int, int> > P;
    for (int i = 0; i < (int)(Q).size(); ++i) {
      int x = Q[i].first, y = Q[i].second;
      for (int d = 0; d < 4; ++d) {
        int nx = x + (d == 2) - (d == 0);
        int ny = y + (d == 3) - (d == 1);
        if (0 <= nx && nx < n && 0 <= ny && ny < m && O[(int)s[nx][ny]] &&
            rk[nx][ny] > rk[x][y]) {
          if (rk[nx][ny] == inf) P.push_back(make_pair(nx, ny));
          rk[nx][ny] = rk[x][y];
          pre[nx][ny] = make_pair(x, y);
        }
      }
    }
    sort((P).begin(), (P).end(), Cmp);
    tmp[0] = ++L;
    for (int i = 1; i < (int)(P).size(); ++i)
      tmp[i] = same(P[i], P[i - 1]) ? tmp[i - 1] : ++L;
    for (int i = 0; i < (int)(P).size(); ++i)
      rk[P[i].first][P[i].second] = tmp[i];
    swap(Q, P);
  }
  for (int i = 0; i < (int)(ok).size(); ++i) O[ok[i]] = false;
  if (rk[ex][ey] == inf) return;
  fd = true;
  string S;
  for (int i = ex, j = ey; i != sx || j != sy;) {
    S += s[i][j];
    int ni = pre[i][j].first, nj = pre[i][j].second;
    i = ni, j = nj;
  }
  reverse((S).begin(), (S).end());
  S.pop_back();
  if ((int)(S).size() < len)
    len = (int)(S).size(), ans = S;
  else if ((int)(S).size() == len)
    ans = min(ans, S);
}
void dfs(int c, int k) {
  if (k > K) return;
  if (c == (int)(has).size()) {
    solve();
    return;
  }
  dfs(c + 1, k);
  ok.push_back(has[c]);
  dfs(c + 1, k + 1);
  ok.pop_back();
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 0; i < n; ++i) scanf("%s", s[i]);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (s[i][j] == 'S')
        sx = i, sy = j;
      else if (s[i][j] == 'T')
        ex = i, ey = j;
      else
        has.push_back(s[i][j]);
    }
  sort((has).begin(), (has).end());
  has.erase(unique((has).begin(), (has).end()), has.end());
  dfs(0, 0);
  if (!fd)
    puts("-1");
  else
    printf("%s\n", ans.c_str());
  return 0;
}
