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 dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int n, m, k, dist[60][60], sx, sy, fx, fy;
string s[60];
bool mark[60][60];
vector<string> v;
string kk;
bool valid(int x, int y) {
  if (x >= 0 && y >= 0 && x < n && y < m) return 1;
  return 0;
}
bool cmp(pair<pair<int, int>, char> x, pair<pair<int, int>, char> y) {
  return x.second < y.second;
}
bool cmp2(string s1, string s2) {
  if (s1.size() < s2.size()) return 1;
  if (s2.size() < s1.size()) return 0;
  return s1 < s2;
}
void dfs(int x, int y, int a, int b, int c, int d) {
  if (x == fx && y == fy) return;
  vector<pair<pair<int, int>, char> > mov;
  for (int i = 0; i < 4; i++) {
    int xx = x + dx[i];
    int yy = y + dy[i];
    if (valid(xx, yy)) {
      if (s[xx][yy] == a + 'a' || s[xx][yy] == b + 'a' ||
          s[xx][yy] == c + 'a' || s[xx][yy] == d + 'a' || s[xx][yy] == 'T') {
        if (dist[xx][yy] != -1 && dist[xx][yy] < dist[x][y]) {
          mov.push_back(pair<pair<int, int>, char>(pair<int, int>(dx[i], dy[i]),
                                                   s[xx][yy]));
        }
      }
    }
  }
  sort(mov.begin(), mov.end(), cmp);
  if (mov.size() > 0) {
    int xx = x + mov[0].first.first;
    int yy = y + mov[0].first.second;
    if (!(xx == fx && yy == fy)) kk += mov[0].second;
    dfs(xx, yy, a, b, c, d);
  }
}
void bfs(int a, int b, int c, int d) {
  memset(mark, 0, sizeof mark);
  memset(dist, -1, sizeof dist);
  queue<pair<int, int> > q;
  q.push(pair<int, int>(fx, fy));
  dist[fx][fy] = 0;
  mark[fx][fy] = 1;
  while (!q.empty()) {
    int x = q.front().first;
    int y = q.front().second;
    q.pop();
    for (int i = 0; i < 4; i++) {
      int xx = x + dx[i];
      int yy = y + dy[i];
      if (valid(xx, yy)) {
        if (s[xx][yy] == a + 'a' || s[xx][yy] == b + 'a' ||
            s[xx][yy] == c + 'a' || s[xx][yy] == d + 'a' || s[xx][yy] == 'S') {
          if (!mark[xx][yy]) {
            mark[xx][yy] = 1;
            dist[xx][yy] = dist[x][y] + 1;
            q.push(pair<int, int>(xx, yy));
          }
        }
      }
    }
  }
  if (dist[sx][sy] == -1) return;
  kk.clear();
  dfs(sx, sy, a, b, c, d);
  v.push_back(kk);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (s[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (s[i][j] == 'T') {
        fx = i;
        fy = j;
      }
    }
  for (int i = 0; i <= 25; i++) {
    if (k > 1) {
      for (int j = i; j <= 25; j++) {
        if (k > 2) {
          for (int z = j; z <= 25; z++) {
            if (k > 3) {
              for (int p = z; p <= 25; p++) {
                bfs(i, j, z, p);
              }
            } else {
              bfs(i, j, z, z);
            }
          }
        } else {
          bfs(i, j, j, j);
        }
      }
    } else {
      bfs(i, i, i, i);
    }
  }
  sort(v.begin(), v.end(), cmp2);
  if (v.size() > 0)
    cout << v[0] << endl;
  else
    cout << -1 << endl;
  return 0;
}
