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;
char tab[64][64];
int pode[64][64];
int DDD[64][64];
map<pair<pair<int, int>, set<char> >, int> dist;
queue<pair<pair<int, int>, set<char> > > fila;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
pair<int, int> INI;
void insere(int d, pair<int, int> p, set<char> s) {
  if (p.first < 0 || p.first >= n || p.second < 0 || p.second >= m) return;
  int tem = s.count(tab[p.first][p.second]);
  if (tab[p.first][p.second] == 'S' && d > 0) return;
  if (!tem && tab[p.first][p.second] >= 'a') {
    if ((int)(s).size() >= k) return;
    s.insert(tab[p.first][p.second]);
  }
  if (!dist.count(make_pair(p, s))) {
    dist.insert(make_pair(make_pair(p, s), d));
    fila.push(make_pair(p, s));
  }
}
queue<pair<pair<int, int>, set<char> > > fila2;
void insere2(int d_ant, pair<int, int> p, set<char> s, int cor, char ANT) {
  if (p.first < 0 || p.first >= n || p.second < 0 || p.second >= m) return;
  if (pode[p.first][p.second] == cor) return;
  int d = 1000000000;
  if (dist.count(make_pair(p, s))) {
    d = dist[make_pair(p, s)];
  }
  if (d_ant - d == 1) {
    fila2.push(make_pair(p, s));
    pode[p.first][p.second] = cor;
    DDD[p.first][p.second] = min(DDD[p.first][p.second], d);
  }
  if (s.count(ANT)) {
    s.erase(ANT);
    d = 1000000000;
    if (dist.count(make_pair(p, s))) {
      d = dist[make_pair(p, s)];
    }
    if (d_ant - d == 1) {
      fila2.push(make_pair(p, s));
      DDD[p.first][p.second] = min(DDD[p.first][p.second], d);
      pode[p.first][p.second] = cor;
    }
  }
}
set<char> SS;
int PODE(int x, int y, int cor, int d_ant) {
  if (x < 0 || x >= n || y < 0 || y >= m) return 0;
  if (pode[x][y] != cor) return 0;
  if (DDD[x][y] == d_ant + 1) return 1;
  return 0;
}
void best_ans(string& atual, set<char>& s, pair<int, int> fim, int cor) {
  SS = s;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++) DDD[i][j] = 1000000000;
  int d = dist[make_pair(fim, s)];
  while (!fila2.empty()) fila2.pop();
  insere2(d + 1, fim, s, cor, '0');
  while (!fila2.empty()) {
    pair<pair<int, int>, set<char> > p = fila2.front();
    fila2.pop();
    d = dist[make_pair(p.first, p.second)];
    char ANT = tab[p.first.first][p.first.second];
    for (int i = 0; i < 4; i++) {
      insere2(d, make_pair(p.first.first + dx[i], p.first.second + dy[i]),
              p.second, cor, ANT);
    }
  }
  pair<int, int> at = INI;
  string nova = "";
  while (at != fim) {
    pair<char, pair<int, int> > vai = make_pair('z' + 1, make_pair(0, 0));
    int dd = DDD[at.first][at.second];
    for (int i = 0; i < (int)(4); i++) {
      if (PODE(at.first + dx[i], at.second + dy[i], cor, dd)) {
        vai =
            min(vai, make_pair(tab[at.first + dx[i]][at.second + dy[i]],
                               make_pair(at.first + dx[i], at.second + dy[i])));
      }
    }
    at = vai.second;
    if (at == fim) break;
    nova += vai.first;
  }
  if ((int)(nova).size() < (int)(atual).size()) atual = nova;
  atual = min(atual, nova);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(n); i++) cin >> tab[i];
  pair<int, int> fim;
  while (!fila.empty()) fila.pop();
  dist.clear();
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++) {
      if (tab[i][j] == 'S') {
        insere(0, make_pair(i, j), set<char>());
        INI = make_pair(i, j);
      }
      if (tab[i][j] == 'T') {
        fim = make_pair(i, j);
      }
    }
  vector<set<char> > ans;
  while (!fila.empty()) {
    pair<pair<int, int>, set<char> > p = fila.front();
    fila.pop();
    int d = dist[p];
    if (tab[p.first.first][p.first.second] == 'T') {
      ans.push_back(p.second);
      continue;
    }
    for (int i = 0; i < 4; i++) {
      insere(d + 1, make_pair(p.first.first + dx[i], p.first.second + dy[i]),
             p.second);
    }
  }
  if ((int)(ans).size() == 0) {
    cout << "-1\n";
  } else {
    memset(pode, -1, sizeof(pode));
    string resp(n * m + 100, 'z');
    for (int i = 0; i < (int)((int)(ans).size()); i++) {
      best_ans(resp, ans[i], fim, i);
    }
    cout << resp << endl;
  }
  return 0;
}
