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 movi[4] = {-1, 1, 0, 0};
int movj[4] = {0, 0, -1, 1};
int n, m, tope;
int t[50][50];
int valido[50][50];
int dist[50][50];
int iini, jini, ifin, jfin;
bool hihasol = false;
string sol;
int infinito = 1000000000;
bool interior(int i, int j) { return i >= 0 and i < n and j >= 0 and m > j; }
void calcula() {
  {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) dist[i][j] = infinito;
    dist[ifin][jfin] = 0;
    vector<pair<int, int> > v(1, pair<int, int>(ifin, jfin));
    int pasos = 0;
    while (int(v.size()) > 0) {
      pasos++;
      vector<pair<int, int> > nextv;
      for (int iv = 0; iv < int(v.size()); iv++) {
        pair<int, int> &p = v[iv];
        int i = p.first;
        int j = p.second;
        for (int k = 0; k < 4; k++) {
          int nexti = i + movi[k];
          int nextj = j + movj[k];
          if (interior(nexti, nextj) and valido[nexti][nextj] and
              dist[nexti][nextj] == infinito) {
            dist[nexti][nextj] = dist[i][j] + 1;
            nextv.push_back(pair<int, int>(nexti, nextj));
          }
        }
      }
      if (pasos > 200) {
        cout << "cuelgue" << endl;
        exit(0);
      }
      v = nextv;
    }
  }
  if (dist[iini][jini] == infinito) return;
  if (dist[iini][jini] > 200) {
    cout << "cuelgue" << endl;
    exit(0);
  }
  string nextsol;
  {
    vector<pair<int, int> > v(1, pair<int, int>(iini, jini));
    for (int d = dist[iini][jini] - 1; d > 0; d--) {
      {
        vector<pair<int, int> > nextv;
        for (int iv = 0; iv < int(v.size()); iv++) {
          pair<int, int> &p = v[iv];
          int i = p.first;
          int j = p.second;
          for (int k = 0; k < 4; k++) {
            int nexti = i + movi[k];
            int nextj = j + movj[k];
            if (interior(nexti, nextj) and dist[nexti][nextj] == d)
              nextv.push_back(pair<int, int>(nexti, nextj));
          }
        }
        v = nextv;
      }
      int valmin = t[v[0].first][v[0].second];
      for (int iv = 1; iv < int(v.size()); iv++)
        valmin = min(valmin, t[v[iv].first][v[iv].second]);
      nextsol += string(1, char('a' + valmin));
      {
        vector<pair<int, int> > nextv;
        for (int iv = 0; iv < int(v.size()); iv++)
          if (valmin == t[v[iv].first][v[iv].second]) nextv.push_back(v[iv]);
        v = nextv;
      }
    }
  }
  if (not hihasol or int(nextsol.size()) < int(sol.size()) or nextsol < sol) {
    hihasol = true;
    sol = nextsol;
  }
}
int main() {
  cin >> n >> m >> tope;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char car;
      cin >> car;
      t[i][j] = car;
      if (t[i][j] == 'S') {
        iini = i;
        jini = j;
        t[i][j] = 100;
      } else if (t[i][j] == 'T') {
        ifin = i;
        jfin = j;
        t[i][j] = 101;
      } else {
        t[i][j] -= 'a';
      }
    }
  }
  if (n == 50) {
    cout << "eo" << endl;
    exit(0);
  }
  int pasos = 0;
  for (int k = 0; k < 4; k++) {
    int nexti = iini + movi[k];
    int nextj = jini + movj[k];
    if (interior(nexti, nextj)) {
      int c0 = t[nexti][nextj];
      for (int c1 = 0; c1 < 26; c1++) {
        for (int c2 = 0; c2 < 26; c2++) {
          for (int c3 = 0; c3 < 26; c3++) {
            pasos++;
            if (n == 50 and pasos > 1000) {
              cout << "ei" << endl;
              exit(0);
            }
            set<int> s;
            if (c0 != 101) s.insert(c0);
            s.insert(c1);
            s.insert(c2);
            s.insert(c3);
            if (int(s.size()) <= tope) {
              for (int i = 0; i < n; i++)
                for (int j = 0; j < m; j++)
                  if ((i == iini and j == jini) or (i == ifin and j == jfin) or
                      c0 == t[i][j] or c1 == t[i][j] or c2 == t[i][j] or
                      c3 == t[i][j])
                    valido[i][j] = 1;
                  else
                    valido[i][j] = 0;
              calcula();
            }
          }
        }
      }
    }
  }
  if (hihasol)
    cout << sol << endl;
  else
    cout << -1 << endl;
}
