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;
int H, W, K;
char m[50][51];
int sx, sy, gx, gy;
bool f[26];
int cs = (1 << 29);
string res = "";
int d[50][50];
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
bool used[50][50];
void solve() {
  for (int(i) = 0; (i) < (H); (i)++)
    for (int(j) = 0; (j) < (W); (j)++) d[i][j] = (1 << 29);
  d[gx][gy] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(gx, gy));
  while (!q.empty()) {
    int x = q.front().first, y = q.front().second;
    q.pop();
    for (int(dd) = 0; (dd) < (4); (dd)++) {
      int nx = x + dx[dd], ny = y + dy[dd];
      if (0 <= nx && nx < H && 0 <= ny && ny < W &&
          (m[nx][ny] == 'S' || m[nx][ny] == 'T' || f[m[nx][ny] - 'a'])) {
        if (d[nx][ny] == (1 << 29)) {
          d[nx][ny] = d[x][y] + 1;
          q.push(make_pair(nx, ny));
        }
      }
    }
  }
  if (d[sx][sy] > cs || d[sx][sy] == (1 << 29)) return;
  string tt = "";
  vector<pair<int, int> > lis;
  memset((used), (0), sizeof(used));
  used[sx][sy] = true;
  lis.push_back(make_pair(sx, sy));
  for (int zan = d[sx][sy]; zan > 1; zan--) {
    vector<pair<int, int> > nls;
    char cgo = '3';
    for (int(i) = 0; (i) < (lis.size()); (i)++) {
      int x = lis[i].first, y = lis[i].second;
      for (int(dd) = 0; (dd) < (4); (dd)++) {
        int nx = x + dx[dd], ny = y + dy[dd];
        if (0 <= nx && nx < H && 0 <= ny && ny < W &&
            d[nx][ny] == d[x][y] - 1) {
          if (cgo == '3')
            cgo = m[nx][ny];
          else
            cgo = min(cgo, m[nx][ny]);
        }
      }
    }
    tt += cgo;
    for (int(i) = 0; (i) < (lis.size()); (i)++) {
      int x = lis[i].first, y = lis[i].second;
      for (int(dd) = 0; (dd) < (4); (dd)++) {
        int nx = x + dx[dd], ny = y + dy[dd];
        if (0 <= nx && nx < H && 0 <= ny && ny < W &&
            d[nx][ny] == d[x][y] - 1 && cgo == m[nx][ny] && !used[nx][ny]) {
          used[nx][ny] = true;
          nls.push_back(make_pair(nx, ny));
        }
      }
    }
    lis.clear();
    lis = nls;
  }
  if (d[sx][sy] == cs)
    res = min(res, tt);
  else
    res = tt;
  cs = min(cs, d[sx][sy]);
}
void dfs(int v, int c) {
  if (v == K) {
    solve();
  } else {
    for (int(i) = (c); (i) < (26); (i)++) {
      f[i] = true;
      dfs(v + 1, i + 1);
      f[i] = false;
    }
  }
}
int main() {
  scanf("%d%d%d", &H, &W, &K);
  for (int(i) = 0; (i) < (H); (i)++) scanf("%s", &m[i]);
  for (int(i) = 0; (i) < (H); (i)++)
    for (int(j) = 0; (j) < (W); (j)++) {
      if (m[i][j] == 'S') sx = i, sy = j;
      if (m[i][j] == 'T') gx = i, gy = j;
    }
  dfs(0, 0);
  if (cs == (1 << 29))
    puts("-1");
  else
    cout << res << endl;
}
