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, X, x, y, a, b;
char A[66][66];
string sol = string(100100, 'z' + 1), scurr;
int P[66];
bool used[66][66];
int tim[66][66];
struct nd {
  int x, y, t;
  nd() {}
  nd(int x, int y, int t) : x(x), y(y), t(t) {}
} curr;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {-1, 0, 1, 0};
bool ok[66][66];
bool usd[66][66];
void DFS(int tx, int ty) {
  if (A[tx][ty] == 'T') {
    ok[tx][ty] = true;
    return;
  }
  usd[tx][ty] = true;
  for (int d = 0; d < 4; d++) {
    int nx = tx + dx[d];
    int ny = ty + dy[d];
    if (tim[nx][ny] == tim[tx][ty] + 1) {
      if (!usd[nx][ny]) DFS(nx, ny);
      ok[tx][ty] |= ok[nx][ny];
    }
  }
}
void BFS() {
  memset(used, false, sizeof(used));
  memset(tim, -1, sizeof(tim));
  queue<nd> Q;
  Q.push(nd(x, y, 0));
  while (!Q.empty()) {
    curr = Q.front();
    Q.pop();
    if (used[curr.x][curr.y]) continue;
    used[curr.x][curr.y] = true;
    tim[curr.x][curr.y] = curr.t;
    for (int d = 0; d < 4; d++) {
      int nx = curr.x + dx[d];
      int ny = curr.y + dy[d];
      if (!used[nx][ny] && (A[nx][ny] == 'T' || P[A[nx][ny] - 'a' + 1]))
        Q.push(nd(nx, ny, curr.t + 1));
    }
  }
  scurr = "";
  if (tim[a][b] == -1) {
    scurr = sol;
    return;
  }
  memset(ok, false, sizeof(ok));
  memset(usd, false, sizeof(usd));
  DFS(x, y);
  int tx = x;
  int ty = y;
  while (A[tx][ty] != 'T') {
    if (A[tx][ty] != 'T' && A[tx][ty] != 'S') scurr += A[tx][ty];
    char best = 'z' + 1;
    int ntx, nty;
    for (int d = 0; d < 4; d++) {
      int nx = tx + dx[d];
      int ny = ty + dy[d];
      if (tim[nx][ny] == 1 + tim[tx][ty] && ok[nx][ny]) {
        if (A[nx][ny] < best) {
          ntx = nx;
          nty = ny;
          best = A[nx][ny];
        }
        if (A[nx][ny] == 'S') {
          ntx = nx;
          nty = ny;
          break;
        }
      }
    }
    tx = ntx;
    ty = nty;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> X;
  for (int I = 1; I <= N; I++) cin >> A[I] + 1;
  for (int I = 1; I <= N; I++)
    for (int K = 1; K <= M; K++) {
      if (A[I][K] == 'S')
        x = I, y = K;
      else if (A[I][K] == 'T')
        a = I, b = K;
    }
  for (int I = 0; I < X; I++) P[26 - I] = 1;
  do {
    BFS();
    if (long long((scurr).size()) < long long((sol).size())) sol = scurr;
    if (long long((scurr).size()) == long long((sol).size()))
      sol = min(sol, scurr);
  } while (next_permutation(P + 1, P + 1 + 26));
  if (long long((sol).size()) == 100100)
    cout << -1 << endl;
  else
    cout << sol << endl;
}
