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>
#pragma comment(linker, "/stack:64000000")
using namespace std;
template <typename first>
inline first abs(const first& a) {
  return a < 0 ? -a : a;
}
template <typename first>
inline first sqr(const first& a) {
  return a * a;
}
const int INF = int(1e9);
const long long INF64 = long long(1e18);
const long double EPS = 1e-9, PI = 3.1415926535897932384626433832795;
const int N = 50 + 3;
int n, m, k;
char a[N][N];
pair<int, int> start, finish;
int ansVal = INF;
string ans(1, 'z' + 1);
int used[256];
int d1[N][N], d2[N][N];
void solve() {
  memset(d1, 63, sizeof(d1));
  queue<pair<int, int> > q;
  d1[start.first][start.second] = 0;
  q.push(start);
  while (!q.empty()) {
    pair<int, int> v = q.front();
    q.pop();
    int x = v.first, y = v.second;
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, -1, 0, 1};
    for (int i = 0; i < int(4); i++) {
      int xx = x + dx[i], yy = y + dy[i];
      if (!(0 <= (xx) && (xx) < (n) && 0 <= (yy) && (yy) < (m))) continue;
      if (!used[a[xx][yy]] && a[xx][yy] != 'S' && a[xx][yy] != 'T') continue;
      if (d1[xx][yy] > d1[x][y] + 1) {
        d1[xx][yy] = d1[x][y] + 1;
        q.push(make_pair((xx), (yy)));
      }
    }
  }
  if (d1[finish.first][finish.second] > INF / 2) return;
  memset(d2, 63, sizeof(d2));
  d2[finish.first][finish.second] = 0;
  q.push(finish);
  while (!q.empty()) {
    pair<int, int> v = q.front();
    q.pop();
    int x = v.first, y = v.second;
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, -1, 0, 1};
    for (int i = 0; i < int(4); i++) {
      int xx = x + dx[i], yy = y + dy[i];
      if (!(0 <= (xx) && (xx) < (n) && 0 <= (yy) && (yy) < (m))) continue;
      if (!used[a[xx][yy]] && a[xx][yy] != 'S' && a[xx][yy] != 'T') continue;
      if (d2[xx][yy] > d2[x][y] + 1) {
        d2[xx][yy] = d2[x][y] + 1;
        q.push(make_pair((xx), (yy)));
      }
    }
  }
  assert(d1[finish.first][finish.second] == d2[start.first][start.second]);
  int ans = d1[finish.first][finish.second];
  if (ansVal < ans) return;
  ansVal = ans;
  string cur;
  vector<pair<int, int> > v;
  v.push_back(start);
  for (int i = 0; i < int(ans - 1); i++) {
    char minc = 'z' + 1;
    vector<pair<int, int> > next;
    for (int j = 0; j < int(int((v).size())); j++) {
      int x = v[j].first, y = v[j].second;
      int dx[] = {-1, 0, 1, 0};
      int dy[] = {0, -1, 0, 1};
      for (int k = 0; k < int(4); k++) {
        int xx = x + dx[k], yy = y + dy[k];
        if (!(0 <= (xx) && (xx) < (n) && 0 <= (yy) && (yy) < (m))) continue;
        if (!used[a[xx][yy]] && a[xx][yy] != 'S' && a[xx][yy] != 'T') continue;
        if (d1[x][y] + 1 + d2[xx][yy] == ans) {
          if (a[xx][yy] > minc) continue;
          if (a[xx][yy] < minc) next.clear();
          minc = a[xx][yy];
          next.push_back(make_pair((xx), (yy)));
        }
      }
    }
    cur.push_back(minc);
    sort((next).begin(), (next).end());
    next.erase(unique((next).begin(), (next).end()), next.end());
    v = next;
  }
  for (int i = 0; i < int(int((cur).size())); i++) assert(islower(cur[i]));
  ::ans = min(::ans, cur);
}
void gen(int idx, int last) {
  if (idx == k) return solve();
  for (int i = last + 1; i < 26; i++)
    if (!used['a' + i]) {
      used['a' + i] = true;
      gen(idx + 1, i);
      used['a' + i] = false;
    }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < int(n); i++) scanf("%s", a[i]);
  for (int i = 0; i < int(n); i++)
    for (int j = 0; j < int(m); j++)
      if (a[i][j] == 'S')
        start = make_pair((i), (j));
      else if (a[i][j] == 'T')
        finish = make_pair((i), (j));
  gen(0, -1);
  if (ansVal > INF / 2)
    puts("-1");
  else
    printf("%s\n", ans.c_str());
  return 0;
}
