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 INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 55;
int n;
int m;
int k;
char M[N][N];
int dist[N][N];
bool reco[N][N];
string SOL;
bool bio[26];
int px, py;
int ex, ey;
bool AA;
inline void load() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%s", M[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (M[i][j] == 'S') px = i, py = j;
      if (M[i][j] == 'T') ex = i, ey = j;
    }
  }
}
queue<pair<int, int> > Q;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
inline bool bfs() {
  memset(dist, -1, sizeof(dist));
  while (!Q.empty()) Q.pop();
  dist[px][py] = 0;
  Q.push(make_pair(px, py));
  while (!Q.empty()) {
    int first = Q.front().first;
    int second = Q.front().second;
    Q.pop();
    for (int i = 0; i < 4; i++) {
      int X = first + dx[i];
      int Y = second + dy[i];
      if (X < 0 || X >= n || Y < 0 || Y >= m || dist[X][Y] != -1) continue;
      if (M[X][Y] != 'T' && !bio[M[X][Y] - 'a']) continue;
      dist[X][Y] = dist[first][second] + 1;
      Q.push(make_pair(X, Y));
    }
  }
  int curr = dist[ex][ey];
  if (curr == -1) return false;
  memset(reco, 0, sizeof(reco));
  Q.push(make_pair(ex, ey));
  reco[ex][ey] = 0;
  while (!Q.empty()) {
    int first = Q.front().first;
    int second = Q.front().second;
    Q.pop();
    for (int i = 0; i < 4; i++) {
      int X = first + dx[i];
      int Y = second + dy[i];
      if (X < 0 || X >= n || Y < 0 || Y >= m ||
          dist[first][second] != dist[X][Y] + 1)
        continue;
      if (reco[X][Y]) continue;
      reco[X][Y] = 1;
      Q.push(make_pair(X, Y));
    }
  }
  Q.push(make_pair(px, py));
  curr = 0;
  string tmp;
  while (!Q.empty()) {
    int first = Q.front().first;
    int second = Q.front().second;
    if (first == ex && second == ey) break;
    Q.pop();
    int mini = (1 << 30);
    bool ok = false;
    for (int i = 0; i < 4; i++) {
      int X = first + dx[i];
      int Y = second + dy[i];
      if (X < 0 || X >= n || Y < 0 || Y >= m || !reco[X][Y]) continue;
      if (dist[first][second] + 1 != dist[X][Y]) continue;
      if (mini == (1 << 30) || M[X][Y] < mini) mini = M[X][Y];
    }
    if (mini != (1 << 30)) tmp += mini;
    for (int i = 0; i < 4; i++) {
      int X = first + dx[i];
      int Y = second + dy[i];
      if (X < 0 || X >= n || Y < 0 || Y >= m || !reco[X][Y]) continue;
      if (dist[first][second] + 1 != dist[X][Y]) continue;
      if (M[X][Y] == mini) Q.push(make_pair(X, Y));
    }
  }
  if (SOL.empty() || tmp < SOL) SOL = tmp;
  AA = true;
  return true;
}
void dfs(int pos, int go) {
  if (pos == k) {
    bfs();
    return;
  }
  for (int i = go + 1; i < 26; i++) {
    bio[i] = 1;
    dfs(pos + 1, i);
    bio[i] = 0;
  }
}
inline void solve() {
  for (int i = -1; i <= 24; i++) {
    memset(bio, 0, sizeof(bio));
    dfs(0, i);
  }
  if (AA) {
    cout << SOL;
  } else
    printf("-1");
  printf("\n");
}
int main(void) {
  load();
  solve();
  return 0;
}
