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 maxn = 55;
const int inf = 1e9 + 100500;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int a[maxn][maxn];
int b[maxn][maxn];
int d[maxn][maxn];
int anc[maxn][maxn];
int qx[maxn * maxn], qy[maxn * maxn];
int l[300];
string best = string(55, 'z');
int n, m, k;
int sx, sy, tx, ty;
void scan();
void go(int, int);
bool dfs(int, int);
void bfs();
void solve();
int main() {
  scan();
  solve();
  return 0;
};
void scan() {
  string s;
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(n); i++) {
    cin >> s;
    for (int j = 0; j < (int)(m); j++) {
      if (s[j] == 'T')
        tx = i, ty = j, a[i][j] = 0;
      else if (s[j] == 'S')
        sx = i, sy = j, a[i][j] = 0;
      else
        a[i][j] = s[j];
    }
  }
};
string cur;
bool dfs(int x, int y) {
  if (x == tx && y == ty) {
    if (best.length() > cur.length() ||
        (best.length() == cur.length() && best > cur))
      best = cur;
    return true;
  }
  b[x][y] = 1;
  {
    vector<pair<int, int> > cand;
    for (int i = 0; i < (int)(4); i++) {
      int cx = x + dx[i], cy = y + dy[i];
      if (((cx) < n && (cx) >= 0 && (cy) < m && (cy) >= 0) && d[cx][cy] != inf)
        cand.push_back(make_pair(a[cx][cy], i));
    }
    sort(cand.begin(), cand.end());
    for (int i = 0; i < (int)(cand.size()); i++) {
      int cx = x + dx[cand[i].second], cy = y + dy[cand[i].second];
      if (b[cx][cy]) continue;
      if (cx != tx || cy != ty) cur.push_back(a[cx][cy]);
      if (dfs(cx, cy)) return true;
      if (cx != tx || cy != ty) cur.resize(cur.size() - 1);
    }
  }
  return false;
};
void go(int pos, int k) {
  if (!k) {
    bfs();
    return;
  }
  for (int i = pos; i <= 'z'; i++) {
    l[i] = 1;
    go(i + 1, k - 1);
    l[i] = 0;
  }
};
void solve() {
  memset(l, 0, sizeof l);
  l[0] = 1;
  go('a', k);
  if (best.length() == 55)
    cout << -1;
  else
    cout << best;
};
void bfs() {
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++) d[i][j] = inf;
  int lq = 0, rq = 1;
  qx[0] = sx, qy[0] = sy;
  d[sx][sy] = 0;
  while (lq != rq) {
    int tx = qx[lq], ty = qy[lq];
    lq++;
    for (int i = 0; i < (int)(4); i++) {
      int cx = tx + dx[i], cy = ty + dy[i];
      if (((cx) < n && (cx) >= 0 && (cy) < m && (cy) >= 0) &&
          d[cx][cy] == inf && l[a[cx][cy]]) {
        d[cx][cy] = d[tx][ty] + 1;
        qx[rq] = cx, qy[rq] = cy;
        rq++;
      }
    }
  }
  if (d[tx][ty] == inf) return;
  for (int i = 0; i < (int)(n); i++) memset(b[i], 0, sizeof b[i]);
  cur = "";
  dfs(sx, sy);
};
