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 <class T>
class Tqueue {
 public:
  Tqueue() {
    h = 0;
    t = 0;
    sz = 0;
  }
  Tqueue(int n) {
    Tqueue();
    resize(n);
  }
  void resize(int n) {
    if (sz > 0) free(data);
    data = (T *)malloc(sizeof(T) * (n + 1));
    sz = n + 1;
    h = t = 0;
  }
  T front() { return data[h]; }
  T back() { return data[t - 1]; }
  int size() {
    int res = t - h;
    if (res < 0) res += sz;
    return res;
  }
  int empty() { return h == t; }
  void pop() {
    h++;
    if (h == sz) h = 0;
  }
  void push(T x) {
    data[t++] = x;
    if (t == sz) t = 0;
  }
  ~Tqueue() {
    if (sz > 0) {
      free(data);
      sz = 0;
    }
  }

 private:
  T *data;
  int h, t, sz;
};
const string name = "c";
const int NMAX = 52;
const int KMAX = 5;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
int n, m, k;
int sx, sy, tx, ty, dist[NMAX][NMAX];
char s[NMAX][NMAX], num[KMAX];
bool may[256];
string ans;
bool good(int x, int y) {
  return (x >= 0 && x < n && y >= 0 && y < m && may[s[x][y]]);
}
vector<pair<int, int> > tmp, now;
int used[NMAX][NMAX];
Tqueue<pair<int, int> > q(NMAX *NMAX);
void solve() {
  memset(may, 0, sizeof(may));
  may['S'] = may['T'] = true;
  for (int i = 0; i < (int)k; i++) may[num[i]] = true;
  q.push(make_pair(tx, ty));
  memset(dist, 255, sizeof(dist));
  dist[tx][ty] = 0;
  while (!q.empty()) {
    int x = q.front().first, y = q.front().second;
    q.pop();
    for (int i = 0; i < (int)4; i++)
      if (dist[x + dx[i]][y + dy[i]] == -1 && good(x + dx[i], y + dy[i])) {
        dist[x + dx[i]][y + dy[i]] = dist[x][y] + 1;
        q.push(make_pair(x + dx[i], y + dy[i]));
      }
  }
  if (dist[sx][sy] == -1) return;
  now.clear();
  now.push_back(make_pair(sx, sy));
  string res = "";
  memset(used, 0, sizeof(used));
  for (int i = dist[sx][sy] - 1; i >= 0; --i) {
    tmp.clear();
    char min_c = num[k - 1];
    for (int j = 0; j < (int)now.size(); j++) {
      int xx, yy, x = now[j].first, y = now[j].second;
      for (int f = 0; f < (int)4; f++) {
        xx = x + dx[f], yy = y + dy[f];
        if (good(xx, yy) && dist[xx][yy] == dist[x][y] - 1)
          min_c = min(min_c, s[xx][yy]);
      }
    }
    for (int j = 0; j < (int)now.size(); j++) {
      int xx, yy, x = now[j].first, y = now[j].second;
      for (int f = 0; f < (int)4; f++) {
        xx = x + dx[f], yy = y + dy[f];
        if (good(xx, yy) && dist[xx][yy] == dist[x][y] - 1 && !used[xx][yy] &&
            s[xx][yy] == min_c) {
          tmp.push_back(make_pair(xx, yy));
          used[xx][yy] = true;
        }
      }
    }
    now = tmp;
    if (i) res += min_c;
  }
  if (ans.length() > res.length()) ans = res;
  if (ans.length() == res.length()) ans = min(ans, res);
}
void gen(int pos, char ch) {
  if (pos == k) {
    solve();
    return;
  }
  if (ch > 'z') return;
  gen(pos, ch + 1);
  num[pos] = ch;
  gen(pos + 1, ch + 1);
}
char c[NMAX][NMAX];
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < (int)n; i++) cin >> s[i];
  for (int i = 0; i < (int)n; i++)
    for (int j = 0; j < (int)m; j++) {
      if (s[i][j] == 'S') sx = i, sy = j;
      if (s[i][j] == 'T') tx = i, ty = j;
    }
  ans = "";
  for (int i = 0; i < (int)n * m + 1; i++) ans += "@";
  gen(0, 'a');
  if (ans[0] == '@')
    cout << -1 << endl;
  else
    cout << ans.data() << endl;
  cerr << clock() << endl;
  return 0;
}