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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
int n, m, k, x, y, xx, yy;
int A[64][64];
string T[64];
vector<int> M;
pair<int, int> P[51][51];
bool B[51][51];
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
int l, r;
void G(int i, int c, int mask, bool ok) {
  if (c > 0 && c <= k && ok) M.push_back(mask);
  if (i < 26) G(i + 1, c, mask, false);
  if (c < k && i < 26) G(i + 1, c + 1, (mask | (1 << i)), true);
}
bool ok(int x, int y) { return (x >= 0 && y >= 0 && x < n && y < m); }
pair<pair<int, char>, pair<pair<int, int>, pair<int, int> > > Q[10000];
void add(int x, int y, int t) {
  r++;
  Q[r] = make_pair(make_pair(t, T[x][y]),
                   make_pair(make_pair(x, y), make_pair(0, 0)));
}
bool comp(pair<pair<int, char>, pair<pair<int, int>, pair<int, int> > > a,
          pair<pair<int, char>, pair<pair<int, int>, pair<int, int> > > b) {
  if (a.first.second < b.first.second) return true;
  if (a.first.second > b.first.second) return false;
  return (a.second.second.first < b.second.second.first);
}
int main() {
  cin >> n >> m >> k;
  for (int(i) = (0); (i) < (n); ++(i)) cin >> T[i];
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (m); ++(j)) {
      if (T[i][j] == 'T') {
        x = i;
        y = j;
      } else if (T[i][j] == 'S') {
        xx = i;
        yy = j;
      }
    }
  for (int(i) = (0); (i) < (n); ++(i))
    for (int(j) = (0); (j) < (m); ++(j))
      if (T[i][j] != 'S' && T[i][j] != 'T') A[i][j] = T[i][j] - 'a';
  G(0, 0, 0, false);
  string res = "-1";
  for (int(i) = (0); (i) < (M.size()); ++(i)) {
    int mask = M[i];
    memset(B, false, sizeof(B));
    for (int(j) = (0); (j) < (n); ++(j))
      for (int(l) = (0); (l) < (m); ++(l)) P[j][l] = make_pair(-1, -1);
    B[x][y] = true;
    l = 0, r = -1;
    add(x, y, 0);
    Q[0].second.second.first = 0;
    int last = -1;
    while (l <= r) {
      pair<int, int> t;
      pair<int, char> q;
      t = Q[l].second.first;
      q = Q[l].first;
      if (q.first != last) {
        int j = 1;
        while (l + j <= r && q.first == Q[l + j].first.first) j++;
        for (int(y) = (l); (y) < (l + j); ++(y)) Q[y].second.second.first = y;
        sort(Q + l, Q + l + j, comp);
      }
      last = q.first;
      t = Q[l].second.first;
      q = Q[l].first;
      l++;
      if (t == make_pair(xx, yy)) break;
      for (int(j) = (0); (j) < (4); ++(j)) {
        int tx = t.first + dx[j], ty = t.second + dy[j];
        if (!ok(tx, ty)) continue;
        if (B[tx][ty]) continue;
        if (T[tx][ty] != 'S' && T[tx][ty] != 'T' &&
            (mask & (1 << A[tx][ty])) == 0)
          continue;
        B[tx][ty] = true;
        P[tx][ty] = t;
        add(tx, ty, q.first + 1);
      }
    }
    pair<int, int> t;
    t = make_pair(xx, yy);
    if (!B[xx][yy]) continue;
    string s = "";
    while (P[t.first][t.second] != make_pair(x, y)) {
      int tx = P[t.first][t.second].first;
      int ty = P[t.first][t.second].second;
      s += T[tx][ty];
      t.first = tx;
      t.second = ty;
    }
    if (res == "-1" || s.size() < res.size() ||
        (s.size() == res.size() && s < res))
      res = s;
  }
  cout << res << endl;
  cin >> n;
  return 0;
}