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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 52;
int n, m, k;
char s[maxN][maxN];
int d[maxN][maxN];
char f[maxN][maxN];
int px[maxN][maxN], py[maxN][maxN];
bool allowed[maxN];
string res;
int length = 1000000;
int sx, sy;
int fx, fy;
bool isOk(int x, int y) {
  return 0 <= x && x < n && 0 <= y && y < m &&
         (s[x][y] == 'S' || s[x][y] == 'T' || allowed[s[x][y] - 'a']);
}
int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
void check() {
  queue<pair<int, int> > q;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      d[i][j] = 1000000000;
      f[i][j] = 'z';
    }
  }
  q.push(make_pair(fx, fy));
  d[fx][fy] = 0;
  while (!q.empty()) {
    pair<int, int> current = q.front();
    q.pop();
    int x = current.first;
    int y = current.second;
    for (int i = 0; i < 4; ++i) {
      int nx = x + dx[i];
      int ny = y + dy[i];
      if (isOk(nx, ny) && d[nx][ny] > d[x][y] + 1) {
        d[nx][ny] = d[x][y] + 1;
        f[nx][ny] = s[x][y];
        px[nx][ny] = x;
        py[nx][ny] = y;
        q.push(make_pair(nx, ny));
      } else if (isOk(nx, ny) && d[nx][ny] == d[x][y] + 1) {
        if (s[x][y] < f[nx][ny]) {
          f[nx][ny] = s[x][y];
          px[nx][ny] = x;
          py[nx][ny] = y;
        }
      }
    }
  }
  if (d[sx][sy] - 1 > length) {
    return;
  }
  string current = "";
  int x = sx;
  int y = sy;
  vector<pair<int, int> > candidates;
  candidates.push_back(make_pair(x, y));
  int len = d[x][y];
  while (len > 1) {
    vector<pair<int, int> > ncandidates;
    char value = 'z';
    for (int i = 0; i < candidates.size(); ++i) {
      int cx = candidates[i].first;
      int cy = candidates[i].second;
      for (int j = 0; j < 4; ++j) {
        int nx = cx + dx[j];
        int ny = cy + dy[j];
        if (isOk(nx, ny) && d[nx][ny] + 1 == d[cx][cy]) {
          if (s[nx][ny] < value) {
            ncandidates.clear();
            ncandidates.push_back(make_pair(nx, ny));
            value = s[nx][ny];
          } else if (s[nx][ny] == value) {
            ncandidates.push_back(make_pair(nx, ny));
          }
        }
      }
    }
    current += value;
    sort(ncandidates.begin(), ncandidates.end());
    candidates = ncandidates;
    candidates.resize(unique(candidates.begin(), candidates.end()) -
                      candidates.begin());
    --len;
  }
  if (current.size() < length || (current.size() == length && current < res)) {
    length = current.size();
    res = current;
  }
}
void rec(int index, int cnt) {
  if (cnt <= k) {
    check();
  }
  if (cnt == k) {
    return;
  }
  for (int i = index; i < 26; ++i) {
    allowed[i] = true;
    rec(i + 1, cnt + 1);
    allowed[i] = false;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    cin >> 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;
      }
      if (s[i][j] == 'T') {
        fx = i;
        fy = j;
      }
    }
  }
  rec(0, 0);
  if (length == 1000000) {
    cout << -1 << endl;
    return 0;
  }
  cout << res << endl;
  return 0;
}
