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;
const int N = 55;
int n, m, k, sx, sy, tx, ty, d[N][N], mp[1 << 8], mrk[N][N];
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, +1, -1};
char str[N][N];
int chk(int x, int y) {
  return x >= 1 && x <= n && y >= 1 && y <= m && mp[str[x][y]];
}
int len = 1e9;
string ans;
void bfs() {
  memset(d, 0x3f, sizeof d);
  d[tx][ty] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(tx, ty));
  while (!q.empty()) {
    int x, y;
    tie(x, y) = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      int ax = x + dx[i], ay = y + dy[i];
      if (chk(ax, ay) && d[ax][ay] > d[x][y] + 1) {
        d[ax][ay] = d[x][y] + 1;
        q.push(make_pair(ax, ay));
      }
    }
  }
  if (d[sx][sy] > n * m) return;
  string s;
  q.push(make_pair(sx, sy));
  while (!q.empty()) {
    char mx = 'z';
    queue<pair<int, int> > qq;
    while (!q.empty()) {
      int x, y;
      tie(x, y) = q.front();
      q.pop();
      for (int i = 0; i < 4; i++) {
        int ax = x + dx[i], ay = y + dy[i];
        if (chk(ax, ay) && d[ax][ay] == d[x][y] - 1) {
          mx = min(mx, str[ax][ay]);
          if (!mrk[ax][ay]) {
            mrk[ax][ay] = 1;
            qq.push(make_pair(ax, ay));
          }
        }
      }
    }
    if (mx >= 'a' && mx <= 'z') s.push_back(mx);
    while (!qq.empty()) {
      int x, y;
      tie(x, y) = qq.front();
      if (str[x][y] == mx) q.push(qq.front());
      mrk[x][y] = 0;
      qq.pop();
    }
  }
  if (d[sx][sy] < len) {
    len = d[sx][sy];
    ans = s;
  } else if (d[sx][sy] == len)
    ans = min(ans, s);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> str[i] + 1;
    for (int j = 1; j <= m; j++) {
      if (str[i][j] == 'S') sx = i, sy = j;
      if (str[i][j] == 'T') tx = i, ty = j;
    }
  }
  mp['S'] = mp['T'] = 1;
  for (int p = 0; p < 26; p++) {
    for (int q = p; q < 26; q++) {
      for (int r = q; r < 26; r++) {
        for (int s = r; s < 26; s++) {
          mp[p + 'a'] = mp[q + 'a'] = mp[r + 'a'] = mp[s + 'a'] = 1;
          int cnt = 1 + (q > p) + (r > q) + (s > r);
          if (cnt <= k) {
            bfs();
          }
          mp[p + 'a'] = mp[q + 'a'] = mp[r + 'a'] = mp[s + 'a'] = 0;
        }
      }
    }
  }
  if (len > n * m)
    cout << -1 << endl;
  else {
    ans.pop_back();
    cout << ans << endl;
  }
}