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;
struct triple {
  int val, r, c, par;
  triple(int _v, int _r, int _c, int _p) { val = _v, r = _r, c = _c, par = _p; }
  bool operator<(const triple &a) const { return val > a.val; }
};
int n, m, k, len;
int rs, cs, re, ce;
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
char mat[60][60];
bool mark[60][60];
int dist[60][60];
int par[60][60];
bool ok[30];
string res;
priority_queue<triple> q[2];
string bfs() {
  for (int i = 0; i < 2; i++)
    while (!q[i].empty()) q[i].pop();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      mark[i][j] = false;
      dist[i][j] = -1;
      par[i][j] = -1;
    }
  }
  dist[rs][cs] = 0;
  int cur = 0, bef = 1, prev, cnt;
  q[cur].push(triple(1, rs, cs, -1));
  while (!q[0].empty() || !q[1].empty()) {
    if (q[bef].empty()) {
      bef = 1 - bef;
      cur = 1 - cur;
      prev = cnt = 0;
    }
    triple t = q[bef].top();
    q[bef].pop();
    int r = t.r;
    int c = t.c;
    int val = t.val;
    if (val > prev) cnt++;
    prev = val;
    if (par[r][c] != -1) continue;
    par[r][c] = t.par;
    if (r == re && c == ce) break;
    for (int i = 0; i < 4; i++) {
      int nr = r + dir[i][0];
      int nc = c + dir[i][1];
      if ((nr == re && nc == ce) ||
          (nr >= 0 && nr < n && nc >= 0 && nc < m && ok[mat[nr][nc] - 'a'])) {
        if (dist[nr][nc] == -1 || dist[nr][nc] == dist[r][c] + 1) {
          dist[nr][nc] = dist[r][c] + 1;
          int nval = cnt * 30 + mat[nr][nc] - 'a';
          q[cur].push(triple(nval, nr, nc, i));
        }
      }
    }
  }
  if (par[re][ce] == -1) return "";
  string res;
  int x = re, y = ce;
  while (x != rs || y != cs) {
    res = mat[x][y] + res;
    int p = par[x][y];
    x -= dir[p][0];
    y -= dir[p][1];
  }
  return res;
}
void bt(int str, int rem) {
  if (rem == 0) {
    string tmp = bfs();
    if (tmp.length() < 1) return;
    tmp = tmp.substr(0, (int)tmp.length() - 1);
    if (len == -1 || tmp.length() < len || (tmp.length() == len && tmp < res)) {
      len = tmp.length();
      res = tmp;
    }
  } else {
    for (int i = str; i < 26; i++) {
      ok[i] = true;
      bt(i + 1, rem - 1);
      ok[i] = false;
    }
  }
}
int main() {
  while (cin >> n >> m >> k) {
    len = -1;
    res = "";
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cin >> mat[i][j];
        if (mat[i][j] == 'S') rs = i, cs = j;
        if (mat[i][j] == 'T') re = i, ce = j;
      }
    }
    bt(0, k);
    if (len == -1)
      cout << -1 << endl;
    else
      cout << res << endl;
  }
  return 0;
}
