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 V = 51;
const int oo = 1000000000;
char map[V][V];
bool in[V], vis[V][V];
int d[V][V];
int tmp[12], sp;
int n, m, K;
struct Node {
  int x, y;
} q[V * V], gd[2][V * V];
int nu[12];
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
string ans;
int sx, sy, ex, ey;
void bfs() {
  int i, j, k, nx, ny, tx, ty, top, tail;
  memset(d, -1, sizeof(d));
  d[ex][ey] = 0;
  top = 0;
  tail = 1;
  q[1].x = ex;
  q[1].y = ey;
  while (top < tail) {
    top++;
    nx = q[top].x;
    ny = q[top].y;
    vis[nx][ny] = 0;
    for (i = 0; i < 4; i++) {
      tx = nx + dx[i];
      ty = ny + dy[i];
      if (tx < 0 || tx >= n) continue;
      if (ty < 0 || ty >= m) continue;
      if (map[tx][ty] != 'S' && !in[map[tx][ty] - 'a']) continue;
      if (d[tx][ty] == -1) {
        d[tx][ty] = d[nx][ny] + 1;
        tail++;
        q[tail].x = tx;
        q[tail].y = ty;
      }
    }
  }
  if (d[sx][sy] == -1) return;
  string tmp = "";
  int fl, te;
  fl = 0;
  nu[fl] = 1;
  gd[0][0].x = sx;
  gd[0][0].y = sy;
  for (i = 1; i < d[sx][sy]; i++) {
    te = 1 - fl;
    nu[te] = 0;
    char ch = 'z';
    for (j = 0; j < nu[fl]; j++) {
      nx = gd[fl][j].x;
      ny = gd[fl][j].y;
      for (k = 0; k < 4; k++) {
        tx = nx + dx[k];
        ty = ny + dy[k];
        if (tx < 0 || tx >= n) continue;
        if (ty < 0 || ty >= m) continue;
        if (d[tx][ty] + 1 != d[nx][ny]) continue;
        if (map[tx][ty] == ch) {
          gd[te][nu[te]].x = tx;
          gd[te][nu[te]++].y = ty;
        } else if (map[tx][ty] < ch) {
          ch = map[tx][ty];
          nu[te] = 0;
          gd[te][nu[te]].x = tx;
          gd[te][nu[te]++].y = ty;
        }
      }
    }
    tmp += ch;
    fl = te;
  }
  if (ans.length() > tmp.length())
    ans = tmp;
  else if (ans.length() < tmp.length())
    return;
  else if (tmp < ans)
    ans = tmp;
}
void dfs(int k, int st) {
  if (k == K) {
    bfs();
    return;
  }
  for (int i = st; i < 26; i++) {
    if (in[i]) continue;
    in[i] = 1;
    tmp[k] = i;
    dfs(k + 1, i + 1);
    in[i] = 0;
  }
}
int i, j, tx, ty;
int main() {
  while (~scanf("%d%d%d", &n, &m, &K)) {
    for (i = 0; i < n; i++) scanf("%s", map[i]);
    memset(in, 0, sizeof(in));
    for (i = 0; i < n; i++)
      for (j = 0; j < m; j++) {
        if (map[i][j] == 'S') {
          sx = i;
          sy = j;
        }
        if (map[i][j] == 'T') {
          ex = i;
          ey = j;
        }
      }
    ans = "";
    for (i = 0; i < n * m + 1; i++) ans += 'z';
    for (i = 0; i < 4; i++) {
      tx = sx + dx[i];
      ty = sy + dy[i];
      if (tx < 0 || tx >= n) continue;
      if (ty < 0 || ty >= m) continue;
      in[map[tx][ty] - 'a'] = 1;
      tmp[0] = map[tx][ty] - 'a';
      dfs(1, 0);
      in[map[tx][ty] - 'a'] = 0;
    }
    if (ans.length() > n * m)
      puts("-1");
    else
      cout << ans << endl;
  }
}
