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 <typename TYPE>
inline TYPE sqr(const TYPE& a) {
  return (a) * (a);
}
const int INF = 1000 * 1000 * 1000;
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 52;
const int AZ = 30;
const int _S = 28;
const int _T = 29;
int n, m, k;
int a[N][N];
bool good[AZ];
bool us[AZ];
pair<int, int> start, finish;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
string ans;
string d[N][N];
void clear_d() {
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) d[i][j] = "";
}
int vt;
pair<int, pair<int, int> > v[4];
void f() {
  queue<pair<int, int> > q;
  q.push(start);
  clear_d();
  while (!q.empty() && d[finish.first][finish.second].length() == 0) {
    pair<int, int> nx = q.front();
    string sx = d[nx.first][nx.second];
    q.pop();
    vt = 0;
    for (int i = 0; i < int(4); ++i) {
      int x = nx.first + dx[i];
      int y = nx.second + dy[i];
      if ((0 <= (x) && (x) < (n) && 0 <= (y) && (y) < (m))) {
        if (good[a[x][y]]) {
          v[vt] = make_pair(a[x][y], pair<int, int>(x, y));
          ++vt;
        } else if (a[x][y] == _T) {
          d[x][y] = sx;
        }
      }
    }
    if (vt > 1) {
      sort(v, v + vt);
    }
    for (int it = 0; it < int(vt); ++it) {
      int x = v[it].second.first;
      int y = v[it].second.second;
      string s = sx;
      s.push_back(char('a' + a[x][y]));
      int dlen = d[x][y].length();
      if (a[x][y] != _S && (dlen == 0 || dlen > s.length() ||
                            (dlen == s.length() && s < d[x][y]))) {
        d[x][y] = s;
        q.push(v[it].second);
      }
    }
  }
  if (d[finish.first][finish.second].length() != 0) {
    string cur = d[finish.first][finish.second];
    if (ans.length() == 0 || ans.length() > cur.length() ||
        (ans.length() == cur.length() && ans > cur)) {
      ans = cur;
    }
  }
}
int main() {
  scanf("%d%d%d\n", &n, &m, &k);
  int az_cnt = 0;
  for (int i = 0; i < int(n); ++i) {
    for (int j = 0; j < int(m); ++j) {
      char c;
      scanf("%c", &c);
      if (c == 'S') {
        a[i][j] = _S;
        start = pair<int, int>(i, j);
      } else if (c == 'T') {
        a[i][j] = _T;
        finish = pair<int, int>(i, j);
      } else {
        a[i][j] = c - 'a';
        if (!us[a[i][j]]) ++az_cnt;
        us[a[i][j]] = true;
      }
    }
    scanf("\n");
  }
  for (int c = 0; c < int(_S); ++c) {
    if (us[c]) {
      good[c] = true;
      if (k < 2 || az_cnt == 1)
        f();
      else
        for (int ch = c + 1; ch < _S; ++ch) {
          if (us[ch] && ch != c) {
            good[ch] = true;
            if (k < 3 || az_cnt == 2)
              f();
            else
              for (int i = ch + 1; i < _S; ++i) {
                if (us[i]) {
                  good[i] = true;
                  if (k < 4 || az_cnt == 3)
                    f();
                  else
                    for (int j = i + 1; j < _S; ++j) {
                      if (us[j]) {
                        good[j] = true;
                        f();
                        good[j] = false;
                      }
                    }
                  good[i] = false;
                }
              }
            good[ch] = false;
          }
        }
      good[c] = false;
    }
  }
  if (ans.length() == 0) {
    puts("-1");
  } else {
    cout << ans << endl;
  }
  return 0;
}
