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;
template <class T>
void pr(T x) {
  cerr << x;
}
template <class T>
void pr(vector<T> v) {
  pr('[');
  for (__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
    pr(it == v.begin() ? "" : ","), pr(*it);
  pr(']');
  ;
}
template <class T1, class T2>
void pr(pair<T1, T2> x) {
  pr(x.first);
  pr(':');
  pr(x.second);
}
template <class T>
void pr(set<T> v) {
  pr('{');
  for (__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
    pr(it == v.begin() ? "" : ","), pr(*it);
  pr('}');
  ;
}
template <class T1, class T2>
void pr(map<T1, T2> v) {
  pr('{');
  for (__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
    pr(it == v.begin() ? "" : ","), pr(*it);
  pr('}');
  ;
}
int N, M, K;
int I0, J0;
char A[100][100];
bool May[300];
const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, -1, 0, 1};
int Q[2][50 * 50 * 4];
int NQ[2];
char Res[1 << 20];
int D[60][60];
int q[50 * 50 * 2 + 10];
int bfs1() {
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < M; ++j) D[i][j] = 1 << 30;
  int head = 0, tail = 0;
  q[tail++] = I0;
  q[tail++] = J0;
  D[I0][J0] = 0;
  while (head != tail) {
    int i = q[head++];
    int j = q[head++];
    for (int k = 0; k < 4; ++k) {
      int ni = i + di[k];
      int nj = j + dj[k];
      if (ni < 0 || ni >= N) continue;
      if (nj < 0 || nj >= M) continue;
      if (A[ni][nj] == 'T') {
        D[ni][nj] = D[i][j] + 1;
        return D[ni][nj];
      }
      if (D[ni][nj] != 1 << 30) continue;
      if (!May[A[ni][nj]]) continue;
      D[ni][nj] = D[i][j] + 1;
      q[tail++] = ni;
      q[tail++] = nj;
    }
  }
  return 1 << 30;
}
string bfs() {
  int nd = bfs1();
  if (nd == 1 << 30) return "-1";
  int n = 0;
  int cur = 0;
  int next = 1;
  NQ[cur] = NQ[next] = 0;
  Q[cur][NQ[cur]++] = I0;
  Q[cur][NQ[cur]++] = J0;
  while (true) {
    char ch = 'z' + 1;
    for (int c = 0; c < NQ[cur]; c += 2) {
      int i = Q[cur][c];
      int j = Q[cur][c + 1];
      for (int k = 0; k < 4; ++k) {
        int ni = i + di[k];
        int nj = j + dj[k];
        if (ni < 0 || ni >= N) continue;
        if (nj < 0 || nj >= M) continue;
        if (D[ni][nj] != D[i][j] + 1) continue;
        if (A[ni][nj] == 'T') {
          Res[n] = 0;
          return string(Res);
        }
        if (!May[A[ni][nj]]) continue;
        ch = min(ch, A[ni][nj]);
      }
    }
    if (ch == 'z' + 1) return "-1";
    Res[n++] = ch;
    NQ[next] = 0;
    for (int c = 0; c < NQ[cur]; c += 2) {
      int i = Q[cur][c];
      int j = Q[cur][c + 1];
      for (int k = 0; k < 4; ++k) {
        int ni = i + di[k];
        int nj = j + dj[k];
        if (ni < 0 || ni >= N) continue;
        if (nj < 0 || nj >= M) continue;
        if (!May[A[ni][nj]]) continue;
        if (ch != A[ni][nj]) continue;
        Q[next][NQ[next]++] = ni;
        Q[next][NQ[next]++] = nj;
      }
    }
    swap(next, cur);
  }
  return "-1";
}
string str;
string ans = "-1";
void rec(int k, int left) {
  if (!left) {
    string r = bfs();
    if (r != "-1" &&
        (ans == "-1" || ((int)((r).size())) < ((int)((ans).size())) ||
         ((int)((r).size())) == ((int)((ans).size())) && r < ans)) {
      ans = r;
    }
  } else {
    for (int i = k; i < ((int)((str).size())); ++i) {
      May[str[i]] = true;
      rec(i + 1, left - 1);
      May[str[i]] = false;
    }
  }
}
int main() {
  cin >> N >> M >> K;
  set<char> s;
  for (int i = 0; i < N; ++i) {
    scanf("%s", A[i]);
    for (int j = 0; j < M; ++j) {
      if (A[i][j] == 'S') {
        I0 = i;
        J0 = j;
      } else if (A[i][j] == 'T') {
      } else
        s.insert(A[i][j]);
    }
  }
  for (set<char>::iterator it = s.begin(); it != s.end(); ++it) str += *it;
  for (int i = 0; i <= K; ++i) {
    rec(0, i);
  }
  cout << ans << endl;
  return 0;
}
