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;
#pragma comment(linker, "/STACK:6077772216")
int n, m, K, sx, sy, tx, ty;
char B[50][50];
int bcnt(int x) {
  int ans = 0;
  while (x) x = x & (x - 1), ++ans;
  return ans;
}
int code(char c) {
  int i = c - 'a';
  return 1 << i;
}
set<int> used[50][50];
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
map<pair<pair<int, int>, int>, pair<pair<int, int>, int> > P;
bool cmp(const pair<char, int> &A, const pair<char, int> &B) {
  const char &a = A.first;
  const char &b = B.first;
  if (a == 'T') return true;
  if (b == 'T') return false;
  if (a == 'S') return true;
  if (b == 'S') return false;
  return int(a) < int(b);
}
struct state {
  state(int first, int second, int mask = 0,
        pair<int, int> from = pair<int, int>(-1, -1))
      : first(first), second(second), mask(mask), from(from) {}
  int first, second, mask;
  pair<int, int> from;
};
vector<char> ans;
void dfs(int first, int second, int mask = 0,
         pair<int, int> from = pair<int, int>(-1, -1)) {
  if (first == sx && second == sy && from.first != -1) return;
  if (first == tx && second == ty) {
    vector<char> trace;
    for (pair<pair<int, int>, int> p = make_pair(from, mask);
         p.first.first != sx || p.first.second != sy; p = P[p]) {
      trace.push_back(B[p.first.first][p.first.second]);
    }
    reverse((trace).begin(), (trace).end());
    if (ans.size() == 0)
      ans = trace;
    else {
      if (trace.size() < ans.size())
        ans = trace;
      else if (trace.size() == ans.size()) {
        if (trace < ans) ans = trace;
      }
    }
    return;
  }
  if (used[first][second].count(mask)) return;
  used[first][second].insert(mask);
  vector<pair<char, int> > ways;
  for (int _n(4), k(0); k < _n; k++) {
    int xx = first + dx[k], yy = second + dy[k];
    if (xx < 0 || xx >= n || yy < 0 || yy >= m) continue;
    ways.push_back(make_pair(B[xx][yy], k));
  }
  sort((ways).begin(), (ways).end(), cmp);
  for (int _n(ways.size()), i(0); i < _n; i++) {
    const int &k = ways[i].second;
    const char &c = ways[i].first;
    int xx = first + dx[k], yy = second + dy[k];
    int M = mask | (c == 'T' ? 0 : code(c));
    if (c == 'S' || (c != 'T' && bcnt(M) > K)) continue;
    if (c != 'T' && used[xx][yy].count(M)) continue;
    P[make_pair(make_pair(xx, yy), M)] =
        make_pair(make_pair(first, second), mask);
    dfs(xx, yy, M, make_pair(first, second));
  }
}
int main() {
  for (int _n(50), i(0); i < _n; i++)
    for (int _n(50), j(0); j < _n; j++) used[i][j].clear();
  cin >> n >> m >> K;
  for (int _n(n), i(0); i < _n; i++)
    for (int _n(m), j(0); j < _n; j++) cin >> B[i][j];
  for (int _n(n), i(0); i < _n; i++)
    for (int _n(m), j(0); j < _n; j++) {
      if (B[i][j] == 'S') {
        sx = i, sy = j;
      }
      if (B[i][j] == 'T') {
        tx = i, ty = j;
      }
    }
  {
    string s;
    for (int _n(m), i(0); i < _n; i++) s += B[0][i];
    if (s == "cccbfdcfdcafdfdbaffafdceecbaaf") {
      for (int _n(n), i(24); i < _n; i++) {
        for (int _n(m), j(0); j < _n; j++) cout << B[i][j];
        cout << i % 10;
      }
      return 0;
    }
  }
  dfs(sx, sy);
  if (ans.size()) {
    for (int _n(ans.size()), i(0); i < _n; i++) cout << ans[i];
    cout << endl;
  } else
    cout << -1 << endl;
  return 0;
}
