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;
int t[50][50];
int infinito = 1000000000;
map<int, int> eleccion[50][50];
int numbits(int c) {
  int num = 0;
  while (c != 0) {
    num++;
    c -= c & -c;
  }
  return num;
}
bool menor(int i1, int j1, int c1, int i2, int j2, int c2) {
  if (i1 == n - 1 and j1 == m - 1) return false;
  if (t[i1][j1] < t[i2][j2]) return true;
  if (t[i1][j1] > t[i2][j2]) return false;
  int nextc1 = c1 | (1 << (t[i1][j1] - 'a'));
  int nextc2 = c2 | (1 << (t[i2][j2] - 'a'));
  if (eleccion[i1][j1][c1])
    i1++;
  else
    j1++;
  if (eleccion[i2][j2][c2])
    i2++;
  else
    j2++;
  return menor(i1, j1, nextc1, i2, j2, nextc2);
}
bool computa(int i, int j, int c) {
  if (i == n - 1 and j == m - 1) return true;
  map<int, int> &e = eleccion[i][j];
  if (e.count(c)) return e[c] != -1;
  int nextc;
  if (i == 0 and j == 0)
    nextc = 0;
  else {
    int car = t[i][j] - 'a';
    if ((c & (1 << car)) == 0 and numbits(c) == k) {
      e[c] = -1;
      return false;
    }
    nextc = c | (1 << car);
  }
  if (i == n - 1) {
    if (computa(i, j + 1, nextc))
      e[c] = 0;
    else
      e[c] = -1;
    return e[c] != -1;
  }
  if (j == m - 1) {
    if (computa(i + 1, j, nextc))
      e[c] = 1;
    else
      e[c] = -1;
    return e[c] != -1;
  }
  if (computa(i, j + 1, nextc)) {
    e[c] = 0;
  } else if (computa(i + 1, j, nextc)) {
    e[c] = 1;
  } else {
    e[c] = -1;
  }
  return e[c] != -1;
}
int main() {
  cin >> n >> m >> k;
  vector<string> v(n);
  int minimoi = infinito, minimoj = infinito, maximoi = 0, maximoj = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    for (int j = 0; j < m; j++) {
      if (v[i][j] == 'S' or v[i][j] == 'T') {
        minimoi = min(minimoi, i);
        maximoi = max(maximoi, i);
        minimoj = min(minimoj, j);
        maximoj = max(maximoj, j);
      }
    }
  }
  vector<vector<int> > nextv(maximoi - minimoi + 1,
                             vector<int>(maximoj - minimoj + 1));
  for (int i = 0; i < int(nextv.size()); i++)
    for (int j = 0; j < int(nextv[i].size()); j++)
      nextv[i][j] = v[minimoi + i][minimoj + j];
  if (nextv[0][0] == 'T') {
    for (int i1 = 0, i2 = int(nextv.size()) - 1; i1 <= i2; i1++, i2--) {
      for (int j1 = 0, j2 = int(nextv[i1].size()); j1 < j2; j1++, j2--) {
        swap(nextv[i1][j1], nextv[i1][j2]);
        swap(nextv[i2][j1], nextv[i2][j2]);
      }
      if (i1 < i2) swap(nextv[i1], nextv[i2]);
    }
  }
  n = int(nextv.size());
  m = int(nextv[0].size());
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) t[i][j] = nextv[i][j];
  if (not computa(0, 0, 0)) {
    cout << -1 << endl;
  } else {
    int i = 0;
    int j = 0;
    int c = 0;
    string sol;
    if (eleccion[i][j][c])
      i++;
    else
      j++;
    while (not(i == n - 1 and j == m - 1)) {
      sol += string(1, char(t[i][j]));
      int nextc = c | (1 << (t[i][j] - 'a'));
      if (eleccion[i][j][c])
        i++;
      else
        j++;
      c = nextc;
    }
    cout << sol << endl;
  }
}
