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;
template <typename S, typename T>
void fill_(S& buf, T x) {
  fill(reinterpret_cast<T*>(&buf),
       reinterpret_cast<T*>(&buf) + sizeof(buf) / sizeof(T), x);
}
template <typename T, int N>
inline int countof(const T (&)[N]) {
  return N;
}
template <typename T>
inline vector<T>& operator+=(vector<T>& ret, const T& x) {
  ret.push_back(x);
  return ret;
}
int begin___ = 0, end___ = 0;
int fifo___[(1000000)];
int originalBoard[(50) + 8][(50) + 8];
int board[(50) + 8][(50) + 8];
int distances[(50) + 8][(50) + 8];
int H, W, K;
int best = (INT_MAX);
int dys[] = {-1, 0, 0, 1};
int dxs[] = {0, -1, 1, 0};
int y0_, x0_, ye_, xe_;
vector<vector<int>> cxs, cys;
string ret;
bool find_pos(int& y, int& x, const int c) {
  for (int i = int(0), CNT_i = int(H); i < CNT_i; ++i)
    for (int j = int(0), CNT_j = int(W); j < CNT_j; ++j)
      if (originalBoard[i][j] == c) {
        y = i;
        x = j;
        return true;
      }
  return false;
}
inline bool in_range(const int y, const int x) {
  return 0 <= y && y < H && 0 <= x && x < W && board[y][x] != '#';
}
bool bfs() {
  fill_(distances, (INT_MAX));
  (begin___ = end___ = 0);
  (fifo___[end___++] =
       (((((0) & 0x3FFF) << 18) | (((ye_)&0x1FF) << 9) | ((xe_)&0x1FF))));
  while (!(begin___ == end___)) {
    int S_cyx = ((fifo___[begin___++])), c = ((S_cyx) >> 18 & 0x3FFF),
        y = ((S_cyx) >> 9 & 0x1FF), x = ((S_cyx)&0x1FF);
    ;
    if (distances[y][x] != (INT_MAX)) continue;
    distances[y][x] = c;
    if (y == y0_ && x == x0_) continue;
    for (int i = int(0), CNT_i = int(countof(dys)); i < CNT_i; ++i) {
      const int y_ = y + dys[i], x_ = x + dxs[i];
      if (!in_range(y_, x_)) continue;
      (fifo___[end___++] =
           (((((c + 1) & 0x3FFF) << 18) | (((y_)&0x1FF) << 9) | ((x_)&0x1FF))));
    }
  }
  return distances[y0_][x0_] != (INT_MAX);
}
void route() {
  const int d = distances[y0_][x0_];
  if (d > best) {
    return;
  } else {
    string r;
    assert(d != (INT_MAX));
    vector<int> xs;
    r.clear();
    (begin___ = end___ = 0);
    (fifo___[end___++] =
         ((((('S') & 0x3FFF) << 18) | (((y0_)&0x1FF) << 9) | ((x0_)&0x1FF))));
    for (int n = d - 1; n > 0; --n) {
      xs.clear();
      while (!(begin___ == end___)) {
        int S_cyx = ((fifo___[begin___++])), c = ((S_cyx) >> 18 & 0x3FFF),
            y = ((S_cyx) >> 9 & 0x1FF), x = ((S_cyx)&0x1FF);
        ;
        for (int i = int(0), CNT_i = int(countof(dys)); i < CNT_i; ++i) {
          const int y_ = y + dys[i], x_ = x + dxs[i];
          if (!in_range(y_, x_) || distances[y_][x_] != n) continue;
          xs += ((((board[y_][x_]) & 0x3FFF) << 18) | (((y_)&0x1FF) << 9) |
                 ((x_)&0x1FF));
        }
      }
      (begin___ = end___ = 0);
      sort((xs).begin(), (xs).end());
      const int L = unique((xs).begin(), (xs).end()) - xs.begin();
      int S_cy_x_ = (xs[0]), c_ = ((S_cy_x_) >> 18 & 0x3FFF),
          y_ = ((S_cy_x_) >> 9 & 0x1FF), x_ = ((S_cy_x_)&0x1FF);
      ;
      for (int i = int(0), CNT_i = int(L); i < CNT_i; ++i) {
        if (((xs[i]) >> 18 & 0x3FFF) == c_)
          (fifo___[end___++] = (xs[i]));
        else
          break;
      }
      r += c_;
    };
    if (d < best) {
      best = d;
      ret = r;
    } else {
      ret = min(ret, r);
    }
  }
}
void dfs(const int k, int next) {
  if (k == 0) {
    if (bfs()) route();
  } else {
    for (int c = next; c <= int('z'); ++c) {
      const vector<int>(&ys) = cys[c], (&xs) = cxs[c];
      for (int i = int(0), CNT_i = int(int((ys).size())); i < CNT_i; ++i)
        board[ys[i]][xs[i]] = c;
      dfs(k - 1, c + 1);
      for (int i = int(0), CNT_i = int(int((ys).size())); i < CNT_i; ++i)
        board[ys[i]][xs[i]] = '#';
    }
  }
}
void init() {
  find_pos(y0_, x0_, 'S');
  find_pos(ye_, xe_, 'T');
  fill_(board, int('#'));
  board[y0_][x0_] = 'S';
  board[ye_][xe_] = 'T';
  fill_(distances, (INT_MAX));
  cys.assign(int('z') + 10, vector<int>());
  cxs.assign(int('z') + 10, vector<int>());
  for (int i = int(0), CNT_i = int(H); i < CNT_i; ++i)
    for (int j = int(0), CNT_j = int(W); j < CNT_j; ++j) {
      cys[originalBoard[i][j]] += i;
      cxs[originalBoard[i][j]] += j;
    }
}
int main(int argc, char* argv[]) {
  string s;
  cin >> H >> W >> K;
  for (int i = int(0), CNT_i = int(H); i < CNT_i; ++i) {
    cin >> s;
    for (int j = int(0), CNT_j = int(int((s).size())); j < CNT_j; ++j)
      originalBoard[i][j] = s[j];
  }
  init();
  dfs(K, 'a');
  if (best == (INT_MAX))
    cout << -1 << endl;
  else
    cout << ret << endl;
  return 0;
}