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:64000000")
using namespace std;
template <typename T>
inline T SQR(T a) {
  return a * a;
}
class __NT {};
template <class T, class U>
struct __TL {};
template <class, class, template <class, bool> class, bool>
struct __IB;
template <class T, class U, class K, template <class, bool> class C, bool B>
struct __IB<T, __TL<U, K>, C, B> : public __IB<T, K, C, B>, public C<U, B> {};
template <class T, class U, template <class, bool> class C, bool B>
struct __IB<T, __TL<T, U>, C, B> : public __IB<T, U, C, B> {};
template <class T, template <class, bool> class C, bool B>
struct __IB<T, __NT, C, B> : public C<T, B> {};
template <typename T, bool b>
struct __IP {
  operator T() const {
    return b ? (std::numeric_limits<T>::has_infinity
                    ? std::numeric_limits<T>::infinity()
                    : std::numeric_limits<T>::max())
             : (std::numeric_limits<T>::lowest());
  }
};
template <bool B = true>
struct __IF
    : public __IB<
          int,
          __IB<long long,
               __IB<double,
                    __IB<long double,
                         __IB<char,
                              __IB<unsigned char,
                                   __IB<bool,
                                        __IB<float,
                                             __IB<unsigned long long, __NT,
                                                  __IP, 0>::__TL<T, __NT>,
                                             __IP, 0>::__TL<T, __NT>,
                                        __IP, 0>::__TL<T, __NT>,
                                   __IP, 0>::__TL<T, __NT>,
                              __IP, 0>::__TL<T, __NT>,
                         __IP, 0>::__TL<T, __NT>,
                    __IP, 0>::__TL<T, __NT>,
               __IP, 0>::__TL<T, __NT>,
          __IP, B> {
  __IF<!B> operator-() const { return __IF<!B>(); }
};
__IF<> INF;
const double PI = 2 * asin(1.0);
void solve();
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
const double eps = 1e-6;
const int maxn = 60;
int Mas[maxn][maxn];
int Used[maxn][maxn];
int n, m, k;
const int mkSz = 'z' - 'a' + 1;
bool allowed[mkSz];
pair<int, int> st, endp;
string ans;
void bfs() {
  queue<pair<int, int> > q;
  memset(Used, 0, sizeof Used);
  q.push(st);
  Used[st.first][st.second] = -1;
  while (!q.empty()) {
    pair<int, int> cur = q.front();
    q.pop();
    if (cur == endp) {
      break;
    }
    int x = cur.first, y = cur.second;
    auto Func = [&](int dx, int dy) {
      if (x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m) {
        int sh = Mas[x + dx][y + dy];
        if (sh == -1 || allowed[sh] == true) {
          if (Used[x + dx][y + dy] == 0) {
            if (dx != 0) Used[x + dx][y + dy] = dx == 1 ? 1 : 2;
            if (dy != 0) Used[x + dx][y + dy] = dy == 1 ? 3 : 4;
            q.push(make_pair(x + dx, y + dy));
          }
        }
      }
    };
    Func(0, 1);
    Func(0, -1);
    Func(1, 0);
    Func(-1, 0);
  }
  if (Used[endp.first][endp.second] == 0) {
    return;
  }
  string Path;
  pair<int, int> pos = endp;
  while (pos != st) {
    Path += Mas[pos.first][pos.second] + 'a';
    int np = Used[pos.first][pos.second];
    if (np == 1) {
      pos.first--;
    } else if (np == 2) {
      pos.first++;
    } else if (np == 3) {
      pos.second--;
    } else if (np == 4) {
      pos.second++;
    } else {
      ++(*(int *)0);
    }
  }
  reverse((Path).begin(), (Path).end());
  if (ans.size() == 0 || int((ans).size()) > int((Path).size())) {
    ans = Path;
  }
  if (int((ans).size()) == int((Path).size())) {
    ans = min(ans, Path);
  }
}
void gen(int k, int l) {
  if (k == 0) {
    bfs();
  } else {
    for (int i = l; i < mkSz - k; ++i) {
      allowed[i] = true;
      gen(k - 1, i + 1);
      allowed[i] = false;
    }
  }
}
void solve() {
  cin >> n >> m >> k;
  string tmp;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> tmp;
    for (int j = 0; j < (int)(m); ++j) {
      if (tmp[j] == 'S') {
        st = make_pair(i, j);
        Mas[i][j] = -1;
      } else if (tmp[j] == 'T') {
        endp = make_pair(i, j);
        Mas[i][j] = -1;
      } else {
        Mas[i][j] = tmp[j] - 'a';
      }
    }
  }
  gen(k, 0);
  if (int((ans).size()) == 0) {
    cout << "-1";
  } else {
    ans.resize(int((ans).size()) - 1);
    cout << ans;
  }
}
