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;
template <class T>
void chmax(T &x, T y) {
  x = x > y ? x : y;
  return;
}
template <class T>
void chmin(T &x, T y) {
  x = x < y ? x : y;
  return;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T gcd(T x, T y) {
  return y == 0 ? x : gcd(y, x % y);
}
template <class T>
T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
const int inf = 1000000000, mod = 1000000007;
const long long infll = 1000000000000000000ll, modll = long long(mod);
const double maxdb = double(infll), eps = 1e-14, PI = acos(-1.0);
template <class T>
inline void read(T &x) {
  T w = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c & 15);
    c = getchar();
  }
  x *= w;
}
template <class T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = ~x + 1;
  if (x) write(x / 10), putchar(x % 10 + '0');
}
const int dx[] = {-1, 0, 0, 1}, dy[] = {0, -1, 1, 0};
const int maxn = 55;
int n, m, k, sx, sy, ex, ey;
int dist[maxn][maxn];
bool choice[205];
char a[maxn][maxn];
vector<char> v[5];
queue<pair<int, int> > q;
pair<int, int> nxt[maxn][maxn];
string ans = "-1";
string getans(int x, int y) {
  if (x == ex && y == ey) return "";
  return a[x][y] + getans(nxt[x][y].first, nxt[x][y].second);
}
void bfs() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) dist[i][j] = inf;
  dist[ex][ey] = 0;
  q.push(make_pair(ex, ey));
  while (!q.empty()) {
    pair<int, int> pos = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      pair<int, int> to = pos;
      to.first += dx[i], to.second += dy[i];
      if (to.first < 1 || to.second < 1 || to.first > n || to.second > m)
        continue;
      if (!choice[a[to.first][to.second]] && a[to.first][to.second] != 'S')
        continue;
      int D = dist[pos.first][pos.second] + 1;
      const pair<int, int> &tmp = nxt[to.first][to.second];
      if (D > dist[to.first][to.second] ||
          D == dist[to.first][to.second] &&
              a[pos.first][pos.second] >= a[tmp.first][tmp.second])
        continue;
      dist[to.first][to.second] = D;
      nxt[to.first][to.second] = pos;
      q.push(to);
    }
  }
  if (dist[sx][sy] == inf) return;
  string res = getans(nxt[sx][sy].first, nxt[sx][sy].second);
  if (ans == "-1")
    ans = res;
  else if ((ans).size() > (res).size())
    ans = res;
  else if ((ans).size() == (res).size() && ans > res)
    ans = res;
  return;
}
void choose(int now) {
  bfs();
  if (now <= k)
    for (int i = 0; i < (v[now]).size(); i++)
      if (!choice[v[now][i]]) {
        choice[v[now][i]] = 1;
        choose(now + 1);
      }
}
void Get(int x, int y, vector<char> &vec) {
  for (int i = 0; i < 4; i++) {
    int tx = x + dx[i], ty = y + dy[i];
    if (tx < 1 || ty < 1 || tx > n || ty > m) continue;
    vec.push_back(a[tx][ty]);
  }
  return;
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    string str;
    cin >> str;
    for (int j = 0; j < m; j++) {
      a[i][j + 1] = str[j];
      if (a[i][j + 1] == 'S')
        sx = i, sy = j + 1;
      else if (a[i][j + 1] == 'T')
        ex = i, ey = j + 1;
    }
  }
  Get(sx, sy, v[1]);
  Get(ex, ey, v[2]);
  for (int i = 0; i < 26; i++) v[3].push_back(i + 'a');
  v[4] = v[3];
  choose(1);
  cout << ans << endl;
  return 0;
}
