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 maxn = 52;
int n, m, k, sx, sy, ex, ey, dx[] = {0, 0, 1, -1}, dy[] = {-1, 1, 0, 0}, mask,
                             cnt[maxn][maxn];
string ans, tmp;
char ar[maxn][maxn], par[maxn][maxn];
bool mark[maxn][maxn];
vector<int> sn, en;
pair<int, int> op, dad[maxn][maxn], x;
bool comp(pair<int, int> a, pair<int, int> b) {
  if (par[a.first][a.second] < par[b.first][b.second])
    return 1;
  else if (par[a.first][a.second] == par[b.first][b.second] &&
           cnt[dad[a.first][a.second].first][dad[a.first][a.second].second] <
               cnt[dad[b.first][b.second].first][dad[b.first][b.second].second])
    return 1;
  return 0;
}
bool check(int x, int y, int i) {
  if (x + dx[i] > -1 && x + dx[i] < n && y + dy[i] > -1 && y + dy[i] < m)
    return 1;
  return 0;
}
void pri(int x, int y) {
  tmp += ar[x][y];
  if (!(ex == x && ey == y)) pri(dad[x][y].first, dad[x][y].second);
}
void dij() {
  vector<pair<int, int>> q1, q2;
  q1.push_back({ex, ey});
  dad[sx][sy] = {-1, -1};
  while (!q1.empty()) {
    int koft = 0;
    while (koft < q1.size()) {
      bool oo = 0;
      x = q1[koft];
      cnt[x.first][x.second] = koft++;
      mark[x.first][x.second] = 1;
      if (ar[x.first][x.second] == 'T' ||
          mask & (1 << ((int)ar[x.first][x.second] - 'a')))
        for (int i = 0; i < 4; i++) {
          int xx = x.first + dx[i], yy = x.second + dy[i];
          if (check(x.first, x.second, i) && !mark[xx][yy]) {
            if (par[xx][yy] == 0)
              par[xx][yy] = ar[x.first][x.second], dad[xx][yy] = x,
              q2.push_back({xx, yy});
            else if (ar[x.first][x.second] < par[xx][yy])
              par[xx][yy] = ar[x.first][x.second], dad[xx][yy] = x;
          }
        }
    }
    q1.clear();
    swap(q1, q2);
    sort(q1.begin(), q1.end(), comp);
  }
  if (dad[sx][sy] != make_pair(-1, -1)) {
    tmp.clear();
    pri(sx, sy);
    if (ans.empty())
      ans = tmp;
    else {
      if (tmp.size() < ans.size())
        ans = tmp;
      else if (tmp.size() == ans.size())
        ans = min(tmp, ans);
    }
  }
  for (int i = 0; i < maxn; i++)
    for (int j = 0; j < maxn; j++) mark[i][j] = 0, par[i][j] = 0;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> ar[i][j];
      if (ar[i][j] == 'S') sx = i, sy = j;
      if (ar[i][j] == 'T') ex = i, ey = j;
    }
  for (int i = 0; i < 4; i++) {
    if (check(sx, sy, i) && ar[sx + dx[i]][sy + dy[i]] != 'T')
      sn.push_back((int)ar[sx + dx[i]][sy + dy[i]] - 'a');
    if (check(ex, ey, i) && ar[ex + dx[i]][ey + dy[i]] != 'S')
      en.push_back((int)ar[ex + dx[i]][ey + dy[i]] - 'a');
  }
  for (int i = 1; i < (1 << 26); i++)
    if (__builtin_popcount(i) <= k) {
      mask = i;
      bool u = 0;
      for (int uu : sn)
        if (mask & (1 << uu)) u = 1;
      if (sn.empty()) u = 1;
      if (u) dij();
    }
  if (ans.empty()) return cout << -1 << endl, 0;
  for (int i = 1; i < ans.size() - 1; i++) cout << ans[i];
  cout << endl;
}
