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 const M = 100 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
int tmp[M][M], ran[M][M], cnt = 0, a[M], indx, indy, ind2x, ind2y;
vector<int> ch2[M];
vector<pair<int, int> > ch;
int dp[M][M], f2;
int c[M][M * M];
pair<int, int> p[M][M];
pair<int, int> q[M * M];
bool mark[M][M];
int n, m, lim, mn = inf, head = 0, tail = 0;
string anss = "";
pair<int, int> now;
void f(int x, int y) {
  bool flag = 0;
  for (int i = 1; i <= 4; i++) {
    if (tmp[x][y] == a[i]) flag = 1;
  }
  if (!flag && (x != 1 || y != 1)) return;
  if (x == 0 || y == 0 || x == n + 1 || y == m + 1) return;
  if (mark[x][y]) return;
  mark[x][y] = 1;
  dp[x][y] = dp[now.first][now.second] + 1;
  p[x][y] = make_pair(now.first, now.second);
  q[head++] = make_pair(x, y);
}
void upd() {
  for (int i = 0; i < cnt; i++) ch2[i].clear();
  for (int i = 0; i < ch.size(); i++) {
    int x = ch[i].first, y = ch[i].second;
    ch2[tmp[x][y]].push_back(ran[p[x][y].first][p[x][y].second]);
  }
  int cnt2 = 0;
  for (int i = 0; i < 26; i++) {
    sort(ch2[i].begin(), ch2[i].end());
    for (int j = 0; j < ch2[i].size(); j++) {
      if (j == 0 || ch2[i][j] != ch2[i][j - 1]) c[i][ch2[i][j]] = cnt2++;
    }
  }
  cnt = cnt2;
  for (int i = 0; i < ch.size(); i++) {
    int x = ch[i].first, y = ch[i].second;
    ran[x][y] = c[tmp[x][y]][ran[p[x][y].first][p[x][y].second]];
  }
}
void c2(int x, int y) {
  if (x == 0 || y == 0 || x == n + 1 || y == m + 1) return;
  if (!dp[x][y]) return;
  if (dp[now.first][now.second] == dp[x][y] + 1 &&
      ran[p[now.first][now.second].first][p[now.first][now.second].second] >
          ran[x][y])
    p[now.first][now.second] = make_pair(x, y);
}
void check() {
  head = 0, tail = 0;
  if (lim == 4 && a[4] == 26) return;
  if (lim != 4 && (a[lim + 1] != 26 || a[lim] == 26)) return;
  for (int i = 1; i <= 4; i++) {
    for (int j = i + 1; j <= 4; j++) {
      if (a[i] == a[j] && a[i] != 26) return;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) dp[i][j] = 0, ran[i][j] = 0, mark[i][j] = 0;
  dp[ind2x][ind2y] = 1;
  int last = 1;
  f2 = 0;
  if (a[1] == 0 && a[2] == 2) f2 = 1;
  q[head++] = make_pair(ind2x, ind2y);
  ch.clear();
  while (head > tail) {
    now = q[tail++];
    if (dp[now.first][now.second] != last) {
      upd();
      ch.clear();
      last++;
    }
    if (tail != 1) {
      c2(now.first - 1, now.second);
      c2(now.first, now.second - 1);
      c2(now.first + 1, now.second);
      c2(now.first, now.second + 1);
    }
    ch.push_back(now);
    f(now.first + 1, now.second);
    f(now.first, now.second + 1);
    f(now.first - 1, now.second);
    f(now.first, now.second - 1);
  }
  if (!dp[indx][indy]) return;
  if (dp[indx][indy] > mn) return;
  int nowx = indx, nowy = indy;
  string ans = "";
  while (true) {
    if (nowx == ind2x && nowy == ind2y) break;
    if (nowx != indx || nowy != indy) ans += char(tmp[nowx][nowy] + 'a');
    int tx = nowx, ty = nowy;
    nowx = p[nowx][nowy].first;
    nowy = p[tx][ty].second;
  }
  if (mn == dp[indx][indy]) {
    if (anss > ans) anss = ans;
  } else
    mn = dp[indx][indy], anss = ans;
}
int main() {
  cin >> n >> m >> lim;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      char r;
      cin >> r;
      if (r == 'S') indx = i, indy = j;
      if (r == 'T') ind2x = i, ind2y = j;
      if (r == 'S') r = 'a';
      if (r == 'T') r = 'a';
      tmp[i][j] = r - 'a';
    }
  }
  for (int i = 0; i < 27; i++) {
    for (int j = i; j < 27; j++) {
      for (int k = j; k < 27; k++) {
        for (int l = k; l < 27; l++) {
          a[1] = i, a[2] = j, a[3] = k, a[4] = l;
          check();
        }
      }
    }
  }
  if (mn == inf)
    cout << -1;
  else
    cout << anss;
}
