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
20 10 3
ebebccacdb
eeebccddeT
eadebecaac
adeeeaccbc
baccccdaed
deabceabba
dadbecbaaa
eacbbcedcb
aeeScdbbab
babaecaead
bacdbebeae
aacbadbeec
acddceecca
acaeaebaba
cdddeaaeae
eabddadade
ddddaeaeed
bccbaacadd
dccccbabdc
ecdaebeccc
 1 30 2
bmjcfldkloleiqqiTnmdjpaSckkijf
 15 15 3
cbbdccabdcbacbd
bcabdcacadacdbc
cbcddbbcdbddcad
ddcabdbbdcabbdc
aabadcccTcabdbb
cbacaaacaabdbbd
dbdcbSdabaadbdb
dbbaddcdddaadbb
bbddcdcbaccbbaa
adadadbdbbddccc
cddbbdaddcbbdcc
bbaadcdbbcaacca
adbdcdbbcbddbcd
cdadbcccddcdbda
cbcdaabdcabccbc
 10 8 2
bdcdcbfa
decffcce
dTffdacb
eeedcdbb
fdbbbcba
ddabfcda
abdbSeed
bdcdcffa
cadbaffa
fcccddad
 20 20 2
ddadfcdeTaeccbedeaec
acafdfdeaffdeabdcefe
abbcbefcdbbbcdebafef
fdafdcccbcdeeaedeffc
dfdaabdefdafabaabcef
febdcabacaaaabfacbbe
abfcaacadfdbfdbaaefd
dacceeccddccaccdbbce
cacebecabedbddfbfdad
dacbfcabbebfddcedffd
cfcdfacfadcfbcebebaa
ddfbebafaccbebeefbac
ebfaebacbbebdfcbcbea
dfbaebcfccacfeaccaad
edeedeceebcbfdbcdbbe
afaacccfbdecebfdabed
ddbdcedacedadeccaeec
ecbSeacbdcccbcedafef
cfdbeeffbeeafccfdddb
cefdbdfbabccfdaaadbf
 2 1 4
S
T
 3 5 2
SbcaT
acbab
acccb
 3 4 1
SbbT
aaaa
abba
 5 3 4
aaT
acc
bbb
bbc
cSb
 1 50 3
SaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaTaaaaaaaaaaa
 3 3 1
aaa
aaa
TSa
 6 6 3
pkhipk
mlfmak
aqmbae
dlbfSj
dpbjcr
aTbqbm
 3 4 1
Sbbb
aaaT
abbc
 5 5 1
caTbc
dccac
dacda
aacaS
cdcab
 1 40 1
faSfgfTcfadcdfagfbccbffbeaaebagbfcfcgdfd
 1 3 3
SaT
 1 10 2
baaSaaTacb
 20 20 2
baaaaaaaaaaaaaaaaaaa
bfffffffacfffffffffa
bgggggggaccgggggggga
bhhhhhhhaccchhhhhhha
biiiiiiiacccciiiiiia
bjjjjjjjacccccjjjjja
bkkkkkkkacccccckkkka
blllllllacccccccllla
bbbbbbbbSccccccccmma
bddddddddddcccccccna
bodddddddcccccccccca
bppddddddddTaaaaaaaa
bqqqdddddddbqqqqqqqa
brrrrddddddbrrrrrrra
bsssssdddddbsssssssa
bttttttddddbttttttta
buuuuuuudddbuuuuuuua
bvvvvvvvvddbvvvvvvva
bwwwwwwwwwdbwwwwwwwa
bbbbbbbbbbbbbbbbbbbb
 10 10 2
baaaaaaaaa
bffacffffa
bggaccggga
bbbSccchha
bdddddccia
bjddccccca
bkkdddTaaa
blllddblla
bmmmmdbmma
bbbbbbbbbb
 15 3 4
llv
ttT
hbo
ogc
kfe
gli
fbx
kfp
spm
cxc
ndw
Soa
pfh
edr
xmv
 3 4 2
Sbbb
aabT
abbc
 5 10 4
aaaaaaaaaa
aaaaaTaaaa
aaaaaaaSaa
aaaaaaaaaa
aaaaaaaaaa
 10 20 3
baaaaaaaaaaaaaaaaaaa
bfffffffacfffffffffa
bgggggggaccgggggggga
bbbbbbbbSccchhhhhhha
biiiiidddddcciiiiiia
bjjjjjjddcccccjjjjja
bkkkkkkkdddTaaaaaaaa
bllllllllddbllllllla
bmmmmmmmmmdbmmmmmmma
bbbbbbbbbbbbbbbbbbbb
 1 2 4
ST
 20 10 4
baaaaaaaaa
bffacffffa
bggaccggga
bhhaccchha
biiaccccia
bjjaccccca
bkkakkkkka
bllallllla
bbbSmmmmma
bnnnnnnnna
booooooooa
bpppppTaaa
bqqqqqbqqa
brrrrrbrra
bdddddbssa
btddddbtta
buudddbuua
bvvvddbvva
bwwwwdbwwa
bbbbbbbbbb
 15 10 4
sejwprqjku
pnjsiopxft
rsplgvwixq
endglkchxl
ftihbbexgh
sxtxbbavge
jcdkusfnmr
skgsqvflia
kcxmcxjpae
amaiwcfile
njgjSunmwd
ldxvahgreu
ecmrajbjuT
naioqigols
pbwrmxkltj
 4 5 3
abaaa
babaT
Sabba
aaaaa
 1 2 1
ST
 2 1 1
S
T
 1 20 3
acbccbbddbffScTadffd

bbbcccaccaac
 -1
 aaca
 bbbbee
 -1
 aacccaa
 bb
 bbbc
 aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
 cbqb
 aaa
 -1
 -1
 a
 aa
 ccccc
 ccccc
 -1
 aab
 aa
 ccccc
 mmmno
 aajbju
 aaba
 c

#include <bits/stdc++.h>
using namespace std;
int n, m, k, is, js, it, jt;
int idir, jdir;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
string Track[55];
struct vertex {
  string value = " ";
  vector<vertex *> neigh = {NULL, NULL, NULL, NULL};
  vertex *parent = NULL;
  vector<pair<int, int>> prev_cells;
};
vertex *Tree = NULL;
int sign(int x) { return (x < 0) ? (-1) : (1); }
bool safe(int x, int y) {
  return (x >= 0 && x < n && y >= 0 && y < m && Track[x][y] != 'S');
}
bool pair_in_vector(vector<pair<int, int>> a, pair<int, int> pr) {
  for (int i = 0; i < a.size(); i++) {
    if (a[i].first == pr.first && a[i].second == pr.second) return true;
  }
  return false;
}
void add_node(int x, int y, vertex *&Tree) {
  if (NULL == Tree) {
    Tree = new vertex;
    Tree->value = "";
    Tree->neigh = {NULL, NULL, NULL, NULL};
    Tree->parent = Tree;
    Tree->prev_cells.push_back({x, y});
  }
  for (int i = 0; i < 4; i++) {
    int row = x + dx[i];
    int col = y + dy[i];
    if (safe(row, col) && !(row == Tree->prev_cells.end()->first &&
                            col == Tree->prev_cells.end()->second)) {
      if (pair_in_vector(Tree->prev_cells, {row, col})) continue;
      Tree->neigh[i] = new vertex;
      Tree->neigh[i]->neigh = {NULL, NULL, NULL, NULL};
      Tree->neigh[i]->value = Tree->value + Track[row][col];
      Tree->neigh[i]->prev_cells = Tree->prev_cells;
      Tree->neigh[i]->prev_cells.push_back({row, col});
      Tree->neigh[i]->parent = Tree;
      if (Track[row][col] == 'T') return;
      add_node(row, col, Tree->neigh[i]);
    }
  }
}
bool need_to_cut(vertex *Tree) {
  set<char> s;
  for (int ind = 0; ind < Tree->value.length(); ind++)
    s.insert(Tree->value[ind]);
  if (s.size() > k + 1)
    return true;
  else
    return false;
}
void cut_tree(vertex *&Tree) {
  if (Tree != NULL) {
    for (int i = 0; i < 4; i++) {
      if (Tree->neigh[i] != NULL && need_to_cut(Tree->neigh[i]))
        Tree->neigh[i] = NULL;
      cut_tree(Tree->neigh[i]);
    }
  }
}
void comparison(vertex *Tree, set<string> &char_set, int &min_length) {
  if (Tree != NULL) {
    for (int i = 0; i < 4; i++) {
      if ((Tree->value[Tree->value.length() - 1] == 'T') &&
          (Tree->value.length() < min_length)) {
        min_length = Tree->value.length();
        string str_copy = Tree->value;
        str_copy.resize(str_copy.size() - 1);
        char_set.insert(str_copy);
      }
      comparison(Tree->neigh[i], *&char_set, *&min_length);
    }
  }
}
int main(int argc, const char *argv[]) {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> Track[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (Track[i][j] == 'S') {
        is = i;
        js = j;
      }
      if (Track[i][j] == 'T') {
        it = i;
        jt = j;
      }
    }
  }
  add_node(is, js, Tree);
  cut_tree(Tree);
  set<string> str_set;
  int min_length = m * n;
  comparison(Tree, str_set, min_length);
  if (!str_set.empty())
    cout << *str_set.begin() << endl;
  else
    cout << "-1";
  return 0;
}
