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;
const int INF = (int)1e9;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int mx[] = {0, 1};
const int my[] = {1, 0};
int n, m, k;
char a[100][100];
inline int sign(int a) {
  if (a == 0) return 0;
  return ((a < 0) ? (-1) : (1));
}
int dx, dy;
vector<char> chrs;
pair<int, int> st, fn, sc;
char tp[100];
pair<int, int> bs, bt;
inline bool good(const pair<int, int>& a) {
  return bs.first <= a.first && a.first < bt.first && bs.second <= a.second &&
         a.second < bt.second;
}
void updateAns(string& ans) {
  string res = "";
  vector<pair<int, int> > cur;
  cur.push_back(sc);
  res.push_back(tp[0]);
  bool madeIt = false;
  while (!cur.empty()) {
    char mn = 'z' + 1;
    for (int i = 0; i < int(int((cur).size())); ++i)
      for (int j = 0; j < int(2); ++j) {
        pair<int, int> nx(cur[i].first + dx * mx[j],
                          cur[i].second + dy * my[j]);
        if (good(nx)) {
          bool ts = false;
          for (int i = 0; i < int(k); ++i)
            if (tp[i] == a[nx.first][nx.second]) {
              ts = true;
              break;
            }
          if (ts || a[nx.first][nx.second] == 'T')
            mn = min(mn, a[nx.first][nx.second]);
        }
      }
    if (mn == 'T') {
      madeIt = true;
      break;
    } else {
      res.push_back(mn);
      vector<pair<int, int> > c2;
      for (int i = 0; i < int(int((cur).size())); ++i)
        for (int j = 0; j < int(2); ++j) {
          pair<int, int> nx(cur[i].first + dx * mx[j],
                            cur[i].second + dy * my[j]);
          if (good(nx) && a[nx.first][nx.second] == mn) {
            c2.push_back(nx);
          }
        }
      sort((c2).begin(), (c2).end());
      c2.erase(unique((c2).begin(), (c2).end()), c2.end());
      cur = c2;
    }
  }
  if (madeIt && (ans == "-1" || ans > res)) ans = res;
}
void solve(int cnt, int idx, string& ans) {
  if (cnt == k) {
    updateAns(ans);
    return;
  }
  for (int i = idx; i < int((chrs).size()); ++i) {
    if (i == tp[0]) continue;
    tp[cnt] = chrs[i];
    solve(cnt + 1, i + 1, ans);
  }
}
int main() {
  cin >> n >> m >> k;
  pair<int, int> st, fn;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == 'S') {
        st = pair<int, int>(i, j);
        continue;
      }
      if (a[i][j] == 'T') {
        fn = pair<int, int>(i, j);
        continue;
      }
      chrs.push_back(a[i][j]);
    }
  sort((chrs).begin(), (chrs).end());
  chrs.erase(unique((chrs).begin(), (chrs).end()), chrs.end());
  bs.first = min(st.first, fn.first);
  bs.second = min(st.second, fn.second);
  bt.first = max(st.first, fn.first) + 1;
  bt.second = max(st.second, fn.second) + 1;
  dx = sign(fn.first - st.first);
  dy = sign(fn.second - st.second);
  if (dx == 0 || dy == 0) {
    string ans = "";
    int cnt = 0;
    pair<int, int> cur = st;
    while (true) {
      cur.first += dx;
      cur.second += dy;
      if (a[cur.first][cur.second] != 'T') {
        if (ans.find(a[cur.first][cur.second]) == string::npos) cnt++;
        ans.push_back(a[cur.first][cur.second]);
      } else {
        break;
      }
    }
    if (cnt > k)
      puts("-1");
    else
      printf("%s\n", ans.c_str());
    return 0;
  }
  string ans = "-1";
  sc = pair<int, int>(st.first + dx, st.second);
  tp[0] = a[sc.first][sc.second];
  solve(1, 0, ans);
  sc = pair<int, int>(st.first, st.second + dy);
  tp[0] = a[sc.first][sc.second];
  solve(1, 0, ans);
  cout << ans << endl;
  return 0;
}
