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;
void out(int res) {
  cout << res << endl;
  exit(0);
}
int n, m, k;
string a[55];
int st, fn;
int di[] = {+1, 0, -1, 0};
int dj[] = {0, +1, 0, -1};
int fl[55][55][5];
int way[55][55][5];
string ss[55][55][5];
bool inBounds(int r, int c) { return (0 <= r && r < n && 0 <= c && c < m); }
string getAns(int R, int C, int k) {
  if (a[R][C] == 'S') return "";
  int p = way[R][C][k];
  int nr = R + di[p];
  int nc = C + dj[p];
  if (a[nr][nc] != a[R][C] && a[R][C] != 'T') k++;
  string res = "";
  if (a[R][C] != 'T') res = a[R][C];
  return getAns(nr, nc, k) + res;
}
void Bfs() {
  set<pair<int, pair<string, int>>> q;
  q.insert(make_pair(0, make_pair("", st * 10 + k)));
  memset(fl, -1, sizeof(fl));
  fl[st / m][st % m][k] = 0;
  ss[st / m][st % m][k] = "0";
  while (!q.empty()) {
    int K, C, R;
    int curr = (*q.begin()).second.second;
    string path = (*q.begin()).second.first;
    q.erase(q.begin());
    K = curr % 10;
    curr /= 10;
    C = curr % m;
    curr /= m;
    R = curr;
    vector<pair<int, int>> v;
    int nr, nc;
    for (int p = (0), _b(4); p < _b; p++)
      if (inBounds(nr = R + di[p], nc = C + dj[p]))
        if (a[nr][nc] != 'S')
          v.push_back(make_pair(a[nr][nc] == 'T' ? -1 : a[nr][nc] - 'a', p));
    sort(v.begin(), v.end());
    for (int i = (0), _b((int)(v).size()); i < _b; i++) {
      nr = R + di[v[i].second];
      nc = C + dj[v[i].second];
      int nk = K;
      if (a[nr][nc] != a[R][C] && a[nr][nc] != 'T') {
        if (!nk) continue;
        nk--;
      }
      string nn = path;
      nn += (v[i].first + 'a');
      if (fl[nr][nc][nk] != -1) {
        if (ss[nr][nc][nk] != "" &&
            ss[nr][nc][nk].substr(1, (int)(ss[nr][nc][nk]).size()) < nn)
          continue;
        ss[nr][nc][nk] = ss[R][C][K];
        ss[nr][nc][nk] += (v[i].first + 'a');
        continue;
      }
      ss[nr][nc][nk] = ss[R][C][K];
      ss[nr][nc][nk] += (v[i].first + 'a');
      fl[nr][nc][nk] = fl[R][C][K] + 1;
      way[nr][nc][nk] = ((v[i].second + 2) & 3);
      if (a[nr][nc] != 'T')
        q.insert(
            make_pair(fl[nr][nc][nk], make_pair(nn, (nr * m + nc) * 10 + nk)));
    }
  }
  int res = -1;
  int Rf = fn / m;
  int Cf = fn % m;
  for (int k = (0), _b(5); k < _b; k++)
    if (fl[Rf][Cf][k] != -1 && (res == -1 || fl[Rf][Cf][k] < res))
      res = fl[Rf][Cf][k];
  if (res == -1) out(res);
  string ans = "";
  for (int k = (0), _b(5); k < _b; k++)
    if (fl[Rf][Cf][k] == res) {
      string curr = ss[Rf][Cf][k].substr(1, (int)(ss[Rf][Cf][k]).size() - 2);
      if (ans == "" || ans > curr) ans = curr;
    }
  cout << ans << endl;
}
int main() {
  cin >> n >> m >> k;
  for (int i = (0), _b(n); i < _b; i++) cin >> a[i];
  for (int i = (0), _b(n); i < _b; i++)
    for (int j = (0), _b(m); j < _b; j++) {
      if (a[i][j] == 'S') st = i * m + j;
      if (a[i][j] == 'T') fn = i * m + j;
    }
  Bfs();
  return 0;
}
