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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
const long long int INF = 2e9;
char ma[51][51];
string dest[51][51];
struct comp {
  bool operator()(const pair<string, pair<int, int> > &f,
                  const pair<string, pair<int, int> > &s) {
    return (
        f.first.length() < s.first.length() ||
        (f.first.length() == s.first.length() &&
         (f.first < s.first || (f.first == s.first && f.second < s.second))));
  }
};
bool compi(string &f, string &s) {
  return (f.length() < s.length() || (f.length() == s.length() && f < s));
}
int n, m, k;
int fx, fy, tx, ty;
string bfs(string &yes) {
  map<char, int> cnt;
  for (int i = 0; i < (yes.length()); i++) {
    cnt[yes[i]]++;
  }
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      dest[i][j] = "";
    }
  set<pair<string, pair<int, int> >, comp> good;
  good.insert(make_pair("", make_pair(fx, fy)));
  const pair<int, int> gone[4] = {make_pair(0, 1), make_pair(0, -1),
                                  make_pair(1, 0), make_pair(-1, 0)};
  while (!good.empty()) {
    string now = good.begin()->first;
    int x = good.begin()->second.first;
    int y = good.begin()->second.second;
    if (x == tx && y == ty) break;
    good.erase(good.begin());
    for (int i = 0; i < (4); i++) {
      int tox = x + gone[i].first;
      int toy = y + gone[i].second;
      if (tox == fx && toy == fy) continue;
      if (x == fx && y == fy && tox == tx && toy == ty) {
        return "";
      }
      if (tox >= 0 && toy >= 0 && tox < n && toy < m) {
        now += ma[tox][toy];
        if ((cnt.count(ma[tox][toy]) || (tox == tx && toy == ty)) &&
            (dest[tox][toy].length() == 0 || compi(now, dest[tox][toy]))) {
          set<pair<string, pair<int, int> >, comp>::iterator it =
              good.find(make_pair(dest[tox][toy], make_pair(tox, toy)));
          if (it != good.end()) good.erase(it);
          dest[tox][toy] = now;
          good.insert(make_pair(dest[tox][toy], make_pair(tox, toy)));
        }
        now.pop_back();
      }
    }
  }
  string out = dest[tx][ty];
  if (out.length() == 0) return "-";
  out.pop_back();
  return out;
}
string minans = "-";
void gen(string &now, int len) {
  if (!len) {
    string with = bfs(now);
    if (with != "-" && (minans == "-" || compi(with, minans))) {
      minans = with;
    }
    return;
  }
  int fr = (!now.length() ? 0 : now.back() - 'a' + 1);
  for (int i = fr; i <= 'z' - 'a'; i++) {
    now += (char)i + 'a';
    gen(now, len - 1);
    now.pop_back();
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      char c;
      cin >> c;
      ma[i][j] = c;
      if (ma[i][j] == 'S') {
        fx = i;
        fy = j;
      }
      if (ma[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  string temp = "";
  gen(temp, k - 1);
  if (minans == "-")
    cout << -1;
  else
    cout << minans;
}
