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>
template <typename T>
T gcd(T a, T b) {
  return (a > b ? (gcd(b, a)) : (a == 0 ? b : (gcd(b % a, a))));
};
template <typename T>
inline T sqr(T a) {
  return a * a;
};
template <typename T>
inline T gmax(T a, T b) {
  return (a > b ? a : b);
};
template <typename T>
inline T gmin(T a, T b) {
  return (a < b ? a : b);
};
using namespace std;
const int maxn = 50 + 5;
const int maxm = 0;
const double esp = 1e-3;
const int BASE = 0;
int i, j;
char a[maxn][maxn], c[maxn][maxn];
string res = "zzzzz";
bool ok = false;
string tmin[maxn][maxn];
vector<string> ss;
int d[maxn][maxn];
int n, m, k;
int sx, sy, tx, ty;
void enter() {
  cin >> n >> m >> k;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      cin >> a[i][j];
      if (a[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (a[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
}
void init() {
  char a, b, c, d;
  string temp;
  if (k >= 4)
    for (a = 'a'; a <= 'z'; a++)
      for (b = a + 1; b <= 'z'; b++)
        for (c = b + 1; c <= 'z'; c++)
          for (d = c + 1; c <= 'z'; d++) {
            temp = "";
            temp += a;
            temp += b;
            temp += c;
            temp += d;
            ss.push_back(temp);
          }
  if (k >= 3)
    for (a = 'a'; a <= 'z'; a++)
      for (b = a + 1; b <= 'z'; b++)
        for (c = b + 1; c <= 'z'; c++) {
          temp = "";
          temp += a;
          temp += b;
          temp += c;
          temp;
          ss.push_back(temp);
        }
  if (k >= 2)
    for (a = 'a'; a <= 'z'; a++)
      for (b = a + 1; b <= 'z'; b++) {
        temp = "";
        temp += a;
        temp += b;
        ss.push_back(temp);
      }
  if (k >= 1)
    for (a = 'a'; a <= 'z'; a++) {
      temp = "";
      temp += a;
      ss.push_back(temp);
    }
  ss.push_back("");
}
int relx[4], rely[4];
inline bool inside(int x, int y) {
  return (x >= 1 && x <= n && y >= 1 && y <= m);
}
inline void process() {
  int i, j;
  memset(d, 127, sizeof(d));
  d[sx][sy] = 0;
  queue<int> qx, qy;
  qx.push(sx);
  qy.push(sy);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) tmin[i][j] = "";
  tmin[sx][sy] = "S";
  int x, y, r, u, v;
  while (!qx.empty()) {
    x = qx.front();
    y = qy.front();
    qx.pop();
    qy.pop();
    for (r = 0; r < 4; r++) {
      u = x + relx[r];
      v = y + rely[r];
      if (!inside(u, v) || c[u][v] == ' ') continue;
      if (d[u][v] < d[x][y] + 1) continue;
      if (d[u][v] > d[x][y] + 1) {
        d[u][v] = d[x][y] + 1;
        tmin[u][v] = tmin[x][y] + c[u][v];
        qx.push(u);
        qy.push(v);
      } else {
        if (tmin[u][v] > tmin[x][y] + c[u][v]) {
          tmin[u][v] = tmin[x][y] + c[u][v];
          qx.push(u);
          qy.push(v);
        }
      }
    }
  }
  if (d[tx][ty] != d[0][0]) {
    ok = true;
    tmin[tx][ty].erase(tmin[tx][ty].length() - 1, 1);
    tmin[tx][ty].erase(0, 1);
    if (res.length() > tmin[tx][ty].length())
      res = tmin[tx][ty];
    else if (res.length() == tmin[tx][ty].length())
      res = gmin(res, tmin[tx][ty]);
  }
}
void solve() {
  int r;
  for (int t = 0; t < ss.size(); t++) {
    for (i = 1; i <= n; i++)
      for (j = 1; j <= m; j++) {
        c[i][j] = ' ';
        if (a[i][j] == 'S' || a[i][j] == 'T') {
          c[i][j] = a[i][j];
          continue;
        }
        for (r = 0; r < ss[t].length(); r++)
          if (a[i][j] == ss[t][r]) {
            c[i][j] = a[i][j];
            break;
          }
      }
    process();
  }
  if (!ok)
    cout << -1;
  else
    cout << res;
}
int main() {
  relx[0] = -1;
  rely[0] = 0;
  relx[1] = 1;
  rely[1] = 0;
  relx[2] = 0;
  rely[2] = -1;
  relx[3] = 0;
  rely[3] = 1;
  enter();
  init();
  solve();
  return 0;
}
