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>
const int magicconst = 73743071;
using namespace std;
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int sx, sy, tx, ty;
int n, m;
int d[55][55];
char u[55][55];
bool b[255];
bool crt(int first, int second) {
  return 0 <= first && first < n && 0 <= second && second < m &&
         b[u[first][second] - 'a'];
}
inline void bfs() {
  queue<pair<int, int> > q;
  q.push(make_pair(tx, ty));
  memset(d, -1, sizeof d);
  d[tx][ty] = 0;
  while (!q.empty()) {
    pair<int, int> temp = q.front();
    q.pop();
    for (int i = 0; i < (int)(4); ++i)
      if (crt(temp.first + dx[i], temp.second + dy[i]) &&
          d[temp.first + dx[i]][temp.second + dy[i]] == -1) {
        d[temp.first + dx[i]][temp.second + dy[i]] =
            d[temp.first][temp.second] + 1;
        q.push(make_pair(temp.first + dx[i], temp.second + dy[i]));
      }
  }
}
char ans[55 * 55];
int t = 0;
int len = -1;
char str[55 * 55];
int w = 0;
pair<int, int> p[55 * 55];
int pt = 0;
pair<int, int> q[55 * 55];
int qt = 0;
int go(int v, int k) {
  if (k == 0) {
    bfs();
    if (d[sx][sy] != -1 && (len == -1 || d[sx][sy] <= len)) {
      w = 0;
      pt = 0;
      qt = 0;
      p[pt++] = make_pair(sx, sy);
      for (int j = 0; j < (int)(d[sx][sy] - 1); ++j) {
        int mm = 100500;
        for (int o = 0; o < (int)(pt); ++o)
          for (int i = 0; i < (int)(4); ++i)
            if (crt(p[o].first + dx[i], p[o].second + dy[i]) &&
                d[p[o].first + dx[i]][p[o].second + dy[i]] ==
                    d[p[o].first][p[o].second] - 1) {
              if (u[p[o].first + dx[i]][p[o].second + dy[i]] < mm) {
                qt = 0;
                q[qt++] = (make_pair(p[o].first + dx[i], p[o].second + dy[i]));
                mm = u[p[o].first + dx[i]][p[o].second + dy[i]];
              } else if (u[p[o].first + dx[i]][p[o].second + dy[i]] == mm) {
                q[qt++] = (make_pair(p[o].first, p[o].second));
              }
            }
        str[w++] = (char)mm;
        pt = 0;
        for (int i = 0; i < (int)(qt); ++i) p[pt++] = (q[i]);
        qt = 0;
      }
      if (len == -1 || len > d[sx][sy]) {
        len = d[sx][sy];
        for (int i = 0; i < (int)(w); ++i) ans[i] = str[i];
      }
      bool ok = false;
      for (int i = 0; i < (int)(w); ++i) {
        if (ans[i] > str[i]) {
          ok = true;
          break;
        } else if (ans[i] < str[i])
          ;
        { break; }
      }
      if (ok) {
        len = d[sx][sy];
        for (int i = 0; i < (int)(len); ++i) ans[i] = str[i];
      }
    }
  } else {
    for (int i = v + 1; i < (int)(26); ++i) {
      b[i] = 1;
      go(i, k - 1);
      b[i] = 0;
    }
  }
}
int main() {
  int k;
  scanf("%d%d%d\n", &n, &m, &k);
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(m); ++j) {
      scanf("%c", &u[i][j]);
      if (u[i][j] == 'S') sx = i, sy = j;
      if (u[i][j] == 'T') tx = i, ty = j;
    }
    scanf("\n");
  }
  memset(b, 0, sizeof b);
  b['S' - 'a'] = 1;
  b['T' - 'a'] = 1;
  go(-1, k);
  if (len == -1)
    printf("-1");
  else {
    for (int i = 0; i < (int)(len); ++i) printf("%c", ans[i]);
  }
  return 0;
}
