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;
char a[55][55];
vector<vector<int> > c;
int was[26];
int wass[55][55];
pair<int, int> Par[55][55];
int x[2510];
int y[2510];
int par[2510];
int d[2510];
int D[55][55];
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
string Path[55][55];
void dfs(int p, int k, vector<int> a) {
  if (k == 0) {
    c.push_back(a);
    return;
  }
  if (p == 26) {
    if (a.size()) c.push_back(a);
    return;
  }
  dfs(p + 1, k, a);
  a.push_back(p);
  dfs(p + 1, k - 1, a);
}
int main() {
  vector<int> aa;
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  dfs(0, k, aa);
  for (int i = (0); i < (n); ++i) scanf("%s", &a[i]);
  int x1, y1, x2, y2;
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == 'S') {
        x1 = i;
        y1 = j;
      } else if (a[i][j] == 'T') {
        x2 = i;
        y2 = j;
      }
  bool f = false;
  string res = "";
  for (int it = (0); it < (c.size()); ++it) {
    memset(wass, 0, sizeof(wass));
    for (int i = (0); i < (n); ++i)
      for (int j = (0); j < (m); ++j) {
        D[i][j] = 2000000000;
        Path[i][j] = "";
      }
    if (it)
      for (int i = (0); i < (c[it - 1].size()); ++i) was[c[it - 1][i]] = 0;
    for (int i = (0); i < (c[it].size()); ++i) was[c[it][i]] = 1;
    int l = 0, r = 0;
    x[0] = x1;
    y[0] = y1;
    d[0] = 0;
    wass[x1][y1] = 1;
    par[0] = -1;
    int v = -1;
    D[x1][y1] = 0;
    while (l <= r) {
      vector<pair<char, pair<int, int> > > can;
      if (a[x[l]][y[l]] == 'T') v = 1;
      for (int k = (0); k < (4); ++k) {
        int xx = x[l] + dx[k];
        int yy = y[l] + dy[k];
        if ((xx > -1) && (yy > -1) && (xx < n) && (yy < m) &&
            ((a[xx][yy] == 'T') || (was[a[xx][yy] - 'a'] == 1))) {
          if (wass[xx][yy] == 0) {
            can.push_back(make_pair(a[xx][yy], make_pair(xx, yy)));
            Par[xx][yy] = make_pair(x[l], y[l]);
          } else if (D[xx][yy] == D[x[l]][y[l]] + 1)
            if (a[Par[xx][yy].first][Par[xx][yy].second] > a[x[l]][y[l]])
              Par[xx][yy] = make_pair(x[l], y[l]);
        }
      }
      sort(can.begin(), can.end());
      for (int i = (0); i < (can.size()); ++i) {
        int xx = can[i].second.first;
        int yy = can[i].second.second;
        wass[xx][yy] = 1;
        r++;
        x[r] = xx;
        y[r] = yy;
        d[r] = d[l] + 1;
        D[xx][yy] = d[r];
        par[r] = l;
      }
      l++;
    }
    if (v != -1) {
      string res1 = "";
      int xx = x2, yy = y2;
      while (1) {
        if (a[xx][yy] == 'S') break;
        res1 += a[xx][yy];
        int xxx = Par[xx][yy].first;
        int yyy = Par[xx][yy].second;
        xx = xxx;
        yy = yyy;
      }
      res1 = res1.substr(1, res1.size() - 1);
      reverse(res1.begin(), res1.end());
      if (!f) {
        f = true;
        res = res1;
      } else if (res1.size() < res.size())
        res = res1;
      else if ((res1.size() == res.size()) && (res1 < res))
        res = res1;
    }
  }
  if (!f)
    printf("-1\n");
  else
    cout << res << endl;
  return 0;
}
