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 n, m, k;
char mp[60][60];
bool can[60][60];
int dist[60][60];
pair<int, int> src, trg;
queue<pair<int, int> > q;
vector<pair<int, int> > pos[300];
int dir[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
int best = 1000000000;
char ans[60 * 60];
char tans[60 * 60];
pair<int, int> posi[2][60 * 60];
int sz[2];
void f(int rem, char st) {
  if (rem == 0) {
    memset(dist, -1, sizeof dist);
    while (!q.empty()) q.pop();
    q.push(pair<int, int>(src.first, src.second));
    dist[src.first][src.second] = 0;
    while (!q.empty()) {
      pair<int, int> tmp = q.front();
      q.pop();
      for (int i = 0; i < 4; i++) {
        int nr = tmp.first + dir[i][0];
        int nc = tmp.second + dir[i][1];
        if (nr >= 0 && nc >= 0 && nr < n && nc < m && can[nr][nc] &&
            dist[nr][nc] == -1) {
          dist[nr][nc] = dist[tmp.first][tmp.second] + 1;
          q.push(pair<int, int>(nr, nc));
        }
      }
    }
    int len = dist[trg.first][trg.second];
    if (len != -1 && len <= best) {
      int cur = 0, nx = 1;
      posi[0][0] = src;
      sz[0] = 1;
      for (int i = 1; i <= len; i++) {
        sz[nx] = 0;
        char mn = 'z' + 1;
        for (int j = 0; j < sz[cur]; j++) {
          for (int k = 0; k < 4; k++) {
            int nr = posi[cur][j].first + dir[k][0];
            int nc = posi[cur][j].second + dir[k][1];
            if (nr >= 0 && nc >= 0 && nr < n && nc < m && dist[nr][nc] == i) {
              mn = min(mn, mp[nr][nc]);
            }
          }
        }
        for (int j = 0; j < sz[cur]; j++) {
          for (int k = 0; k < 4; k++) {
            int nr = posi[cur][j].first + dir[k][0];
            int nc = posi[cur][j].second + dir[k][1];
            if (nr >= 0 && nc >= 0 && nr < n && nc < m && dist[nr][nc] == i &&
                mp[nr][nc] == mn) {
              posi[nx][sz[nx]++] = pair<int, int>(nr, nc);
            }
          }
        }
        tans[i - 1] = mn;
        cur = 1 - cur, nx = 1 - nx;
      }
      bool good = true;
      if (best == len) {
        for (int i = 0; i < len; i++)
          if (tans[i] != ans[i]) {
            if (tans[i] < ans[i])
              break;
            else
              good = false;
          }
      }
      if (good) {
        for (int i = 0; i < len; i++) ans[i] = tans[i];
        best = len;
      }
    }
    return;
  }
  for (char i = st; i <= 'z'; i++) {
    for (int j = 0; j < pos[i].size(); j++)
      can[pos[i][j].first][pos[i][j].second] = true;
    f(rem - 1, i + 1);
    for (int j = 0; j < pos[i].size(); j++)
      can[pos[i][j].first][pos[i][j].second] = false;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mp[i][j];
      if (mp[i][j] == 'S')
        src = pair<int, int>(i, j);
      else if (mp[i][j] == 'T')
        trg = pair<int, int>(i, j);
      else
        pos[mp[i][j]].push_back(pair<int, int>(i, j));
    }
  }
  can[trg.first][trg.second] = true;
  f(k, 'a');
  if (best == 1000000000)
    cout << -1 << endl;
  else {
    ans[best - 1] = 0;
    puts(ans);
  }
  return 0;
}
