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 MAX = 60;
const int A = 26;
const int INF = 1000000001;
char G[MAX + 2][MAX + 3];
int D[MAX + 2][MAX + 2];
bool ok[A + 1];
pair<int, int> S, T;
vector<char> ans;
vector<int> K;
int n, m, k;
inline int con(char c) { return 'a' <= c && c <= 'z' ? c - 'a' : A; }
vector<pair<int, int> > Q;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
bool vis[A + 1][A + 1][A + 1][A + 1];
void bfs() {
  sort((K).begin(), (K).end());
  K.erase(unique((K).begin(), (K).end()), K.end());
  if (K.back() == A) K.pop_back();
  if (K.size() > k || K.empty()) return;
  if (vis[K[0]][K.size() > 1 ? K[1] : A][K.size() > 2 ? K[2] : A]
         [K.size() > 3 ? K[3] : A])
    return;
  vis[K[0]][K.size() > 1 ? K[1] : A][K.size() > 2 ? K[2] : A]
     [K.size() > 3 ? K[3] : A] = true;
  for (int i = 0; i < K.size(); ++i) ok[K[i]] = true;
  for (int i = 0; i <= n + 1; ++i) memset(D[i], -1, sizeof(int) * (m + 2));
  Q.clear();
  Q.push_back(T);
  D[T.first][T.second] = 0;
  for (int foo = 0; foo < Q.size(); ++foo) {
    pair<int, int> v = Q[foo];
    if (v == S) break;
    for (int foo = 0; foo < 4; ++foo) {
      int i = v.first + dx[foo];
      int j = v.second + dy[foo];
      if (ok[con(G[i][j])] && D[i][j] == -1) {
        D[i][j] = D[v.first][v.second] + 1;
        Q.push_back(make_pair(i, j));
      }
    }
  }
  for (int i = 0; i < K.size(); ++i) ok[K[i]] = false;
  int d = INF;
  for (int foo = 0; foo < 4; ++foo) {
    int i = S.first + dx[foo];
    int j = S.second + dy[foo];
    if (D[i][j] != -1) d = min(d, D[i][j]);
  }
  if (d > ans.size()) return;
  int com = 0;
  int index = 0;
  vector<char> tmp;
  tmp.reserve(d);
  vector<pair<int, int> > V[2];
  bool now = false;
  V[0].push_back(S);
  while (d) {
    int best = 200;
    for (int i = 0; i < K.size(); ++i)
      for (__typeof((V[now]).begin()) u = (V[now]).begin(); u != (V[now]).end();
           ++u)
        for (int foo = 0; foo < 4; ++foo) {
          int x = u->first + dx[foo];
          int y = u->second + dy[foo];
          if (G[x][y] == K[i] + 'a' && D[x][y] == d) {
            best = K[i];
            goto out;
          }
        }
  out:
    tmp.push_back(best + 'a');
    if (!com) {
      if (tmp.back() < ans[index])
        com = -1;
      else if (tmp.back() > ans[index])
        com = 1;
    }
    ++index;
    V[!now].clear();
    for (__typeof((V[now]).begin()) u = (V[now]).begin(); u != (V[now]).end();
         ++u)
      for (int foo = 0; foo < 4; ++foo) {
        int x = u->first + dx[foo];
        int y = u->second + dy[foo];
        if (G[x][y] == best + 'a' && D[x][y] == d) {
          V[!now].push_back(make_pair(x, y));
          D[x][y] = -1;
        }
      }
    --d;
    now = !now;
  }
  if (tmp.size() < ans.size() || com == -1) ans = tmp;
}
void solve(int x, int i) {
  if (x >= k) {
    vector<int> tmpK = K;
    bfs();
    K = tmpK;
    return;
  }
  if (i == A) return;
  solve(x, i + 1);
  for (; i < A; ++i) {
    K.push_back(i);
    solve(x + 1, i + 1);
    K.pop_back();
  }
}
int main(int argc, char *argv[]) {
  scanf("%d %d %d", &n, &m, &k);
  ans.resize(n * m + 10);
  for (int i = 0; i <= n + 1; ++i) memset(G[i], '$', m + 2);
  for (int i = 1; i <= n; ++i) scanf("%s", G[i] + 1);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (G[i][j] == 'S')
        S = make_pair(i, j);
      else if (G[i][j] == 'T')
        T = make_pair(i, j);
  if (abs(S.first - T.first) + abs(S.second - T.second) == 1) return 0;
  for (int xs = S.first - 1; xs <= S.first + 1; ++xs)
    for (int ys = S.second - 1; ys <= S.second + 1; ++ys)
      if (abs(xs - S.first) + abs(ys - S.second) == 1)
        for (int xt = T.first - 1; xt <= T.first + 1; ++xt)
          for (int yt = T.second - 1; yt <= T.second + 1; ++yt)
            if (abs(xt - T.first) + abs(yt - T.second) == 1) {
              K.clear();
              K.push_back(con(G[xs][ys]));
              K.push_back(con(G[xt][yt]));
              if (K[0] == K[1]) K.pop_back();
              solve(K.size(), 0);
            }
  if (ans.size() == n * m + 10) {
    printf("-1\n");
    return 0;
  }
  for (__typeof((ans).begin()) i = (ans).begin(); i != (ans).end(); ++i)
    printf("%c", *i);
  printf("\n");
  return 0;
}
