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;
int sx, sy;
int fx, fy;
int n, m, k;
char buf[55][55];
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int tm[55][55][5];
int type[55][55][5];
int prv[55][55][5][3];
struct pos {
  int x, y, mask;
};
int count(int mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) {
      ret++;
    }
    mask >>= 1;
  }
  return ret;
}
int kol = 0;
int q = 0;
int arr[5000];
map<int, int> tt;
bool BFS() {
  queue<pos> Q;
  pos cur = {sx, sy, 0};
  tm[sx][sy][0] = 0;
  type[sx][sy][0] = 0;
  int pt = 0;
  tt[0] = 1;
  prv[sx][sy][0][0] = -2;
  Q.push(cur);
  while (!Q.empty()) {
    cur = Q.front();
    int kk = count(cur.mask);
    if (tm[cur.x][cur.y][kk] != pt) {
      queue<pos> Q2;
      kol = 0;
      q = 1;
      tt.clear();
      while (!Q.empty()) {
        pos tec = Q.front();
        Q.pop();
        arr[kol++] = type[tec.x][tec.y][count(tec.mask)];
        Q2.push(tec);
      }
      Q = Q2;
      sort(arr, arr + kol);
      tt[arr[0]] = q++;
      for (int i = 1; i < kol; i++) {
        if (arr[i] != arr[i - 1]) {
          tt[arr[i]] = q++;
        }
      }
      pt = tm[cur.x][cur.y][kk];
      continue;
    }
    Q.pop();
    for (int i = 0; i < 4; i++) {
      int nx = cur.x + dx[i];
      int ny = cur.y + dy[i];
      if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
        if (buf[nx][ny] == 'T') {
          if (tm[nx][ny][kk] == -1) {
            tm[nx][ny][kk] = tm[cur.x][cur.y][kk] + 1;
            prv[nx][ny][kk][0] = cur.x;
            prv[nx][ny][kk][1] = cur.y;
            prv[nx][ny][kk][2] = kk;
            type[nx][ny][kk] = tt[type[cur.x][cur.y][kk]] * 5000;
          } else if (tm[nx][ny][kk] == tm[cur.x][cur.y][kk] + 1 &&
                     type[nx][ny][kk] > tt[type[cur.x][cur.y][kk]] * 5000) {
            tm[nx][ny][kk] = tm[cur.x][cur.y][kk] + 1;
            prv[nx][ny][kk][0] = cur.x;
            prv[nx][ny][kk][1] = cur.y;
            prv[nx][ny][kk][2] = kk;
            type[nx][ny][kk] = tt[type[cur.x][cur.y][kk]] * 5000;
          }
        } else {
          int nmask = cur.mask | (1 << (buf[nx][ny] - 'a'));
          if (count(nmask) <= k) {
            int kkk = count(nmask);
            if (tm[nx][ny][kkk] == -1) {
              tm[nx][ny][kkk] = tm[cur.x][cur.y][kk] + 1;
              prv[nx][ny][kkk][0] = cur.x;
              prv[nx][ny][kkk][1] = cur.y;
              prv[nx][ny][kkk][2] = kk;
              type[nx][ny][kkk] =
                  tt[type[cur.x][cur.y][kk]] * 5000 + buf[nx][ny] - 'a';
              pos nxt = {nx, ny, nmask};
              Q.push(nxt);
            } else if (tm[nx][ny][kkk] == tm[cur.x][cur.y][kk] + 1 &&
                       type[nx][ny][kkk] > tt[type[cur.x][cur.y][kk]] * 5000 +
                                               buf[nx][ny] - 'a') {
              tm[nx][ny][kkk] = tm[cur.x][cur.y][kk] + 1;
              prv[nx][ny][kkk][0] = cur.x;
              prv[nx][ny][kkk][1] = cur.y;
              prv[nx][ny][kkk][2] = kk;
              type[nx][ny][kkk] =
                  tt[type[cur.x][cur.y][kk]] * 5000 + buf[nx][ny] - 'a';
              pos nxt = {nx, ny, nmask};
              Q.push(nxt);
            }
          }
        }
      }
    }
  }
  for (int i = 0; i <= k; i++) {
    if (tm[fx][fy][i] != -1) {
      return true;
    }
  }
  return false;
}
void DFS(int px, int py, int pc) {
  if (prv[px][py][pc][0] == -2) {
    return;
  }
  DFS(prv[px][py][pc][0], prv[px][py][pc][1], prv[px][py][pc][2]);
  if (px != fx || py != fy) {
    printf("%c", buf[px][py]);
  }
}
int main() {
  memset(tm, -1, sizeof tm);
  memset(prv, -1, sizeof prv);
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> buf[i];
    for (int j = 0; j < m; j++) {
      if (buf[i][j] == 'S') {
        sx = i;
        sy = j;
      } else if (buf[i][j] == 'T') {
        fx = i;
        fy = j;
      }
    }
  }
  if (BFS()) {
    int bk = 0, bt = 1 << 30, bv = 0;
    for (int i = 0; i <= k; i++) {
      if (tm[fx][fy][i] != -1 &&
          (tm[fx][fy][i] < bt ||
           (tm[fx][fy][i] == bt && type[fx][fy][i] < bv))) {
        bk = i;
        bt = tm[fx][fy][i];
        bv = type[fx][fy][i];
      }
    }
    DFS(fx, fy, bk);
    cout << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}
