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 N = 55;
int x1, y11, x2, y2, n, m, d[N][N], p[26], mrk[N][N];
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, +1, -1};
char str[N][N];
int chk(char ch) { return (ch == 'S' || ch == 'T' || p[ch - 'a']); }
string tmp;
int bfs() {
  memset(d, 0x3f, sizeof d);
  d[x2][y2] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(x2, y2));
  while (!q.empty()) {
    int x, y;
    tie(x, y) = q.front();
    q.pop();
    for (int j = 0; j < 4; j++) {
      int ax = x + dx[j], ay = y + dy[j];
      if (ax >= 1 && ax <= n && ay >= 1 && ay <= m && chk(str[ax][ay]) &&
          d[ax][ay] > d[x][y]) {
        d[ax][ay] = d[x][y] + 1;
        q.push(make_pair(ax, ay));
      }
    }
  }
  if (d[x1][y11] > n * m) return 0;
  tmp.clear();
  q.push({x1, y11});
  while (!q.empty()) {
    int mx = 26;
    queue<pair<int, int> > v;
    while (q.size()) {
      int x, y;
      tie(x, y) = q.front();
      q.pop();
      for (int i = 0; i < 4; i++) {
        int ax = x + dx[i], ay = y + dy[i];
        if (ax >= 1 && ax <= n && ay >= 1 && ay <= m &&
            d[ax][ay] == d[x][y] - 1) {
          if (mx >= str[ax][ay] - 'a') {
            mx = str[ax][ay] - 'a';
            if (!mrk[ax][ay]) v.push(make_pair(ax, ay));
            mrk[ax][ay] = 1;
          }
        }
      }
    }
    if (mx >= 0 && mx < 26) tmp += (char)(mx + 'a');
    while (v.size()) {
      int x, y;
      tie(x, y) = v.front();
      v.pop();
      if (str[x][y] == mx + 'a') q.push(make_pair(x, y));
      mrk[x][y] = 1;
    }
  }
  return 1;
}
int32_t main() {
  int kk;
  cin >> n >> m >> kk;
  for (int i = 1; i <= n; i++) {
    cin >> str[i] + 1;
    for (int j = 1; j <= m; j++) {
      if (str[i][j] == 'S') x1 = i, y11 = j;
      if (str[i][j] == 'T') x2 = i, y2 = j;
    }
  }
  int ans = 1e9;
  string a;
  for (int i = 0; i < 26; i++) {
    int mask = (1 << i);
    p[i] = 1;
    if (kk == 1 && bfs()) {
      if (d[x1][y11] < ans) {
        ans = d[x1][y11];
        a = tmp;
      } else if (ans == d[x1][y11])
        a = min(a, tmp);
    }
    for (int j = i + 1; j < 26; j++) {
      p[j] = 1;
      mask |= (1 << j);
      if (kk == 2 && bfs()) {
        if (d[x1][y11] < ans) {
          ans = d[x1][y11];
          a = tmp;
        } else if (ans == d[x1][y11])
          a = min(a, tmp);
      }
      for (int k = j + 1; k < 26; k++) {
        p[k] = 1;
        mask |= (1 << k);
        if (kk == 3 && bfs()) {
          if (d[x1][y11] < ans) {
            ans = d[x1][y11];
            a = tmp;
          } else if (ans == d[x1][y11])
            a = min(a, tmp);
        }
        for (int l = k + 1; l < 26; l++) {
          p[l] = 1;
          mask |= (1 << l);
          if (kk == 4 && bfs()) {
            if (d[x1][y11] < ans) {
              ans = d[x1][y11];
              a = tmp;
            } else if (ans == d[x1][y11])
              a = min(a, tmp);
          }
          p[l] = 0;
        }
        p[k] = 0;
      }
      p[j] = 0;
    }
    p[i] = 0;
  }
  if (ans > n * m)
    cout << -1 << endl;
  else
    cout << a << endl;
}
