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;
const int dir[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
struct Node {
  int x, y;
  Node() {}
  Node(int a, int b) : x(a), y(b) {}
} S, T, pre[55][55];
int n, m, k, l;
int res = 100000000;
int dis[55][55];
string way;
string rec;
char maps[55][55];
queue<Node> que;
void bfs(string ste) {
  int i, j;
  Node u, t;
  string tway[55][55];
  bool ok = true;
  while (!que.empty()) que.pop();
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      dis[i][j] = 100000000;
    }
  dis[S.x][S.y] = 0;
  tway[S.x][S.y] = "";
  que.push(S);
  while (!que.empty()) {
    int cnt = 0;
    Node next[55 * 55];
    while (!que.empty()) {
      u = que.front();
      que.pop();
      if (u.x == T.x && u.y == T.y) {
        ok = false;
        break;
      }
      for (int i = 0; i < 4; i++) {
        t = u;
        t.x += dir[i][0];
        t.y += dir[i][1];
        if (t.x >= 0 && t.x < n && t.y >= 0 && t.y < m) {
          j = ste.find(maps[t.x][t.y]);
          if (j == -1 && maps[t.x][t.y] != 'T') continue;
          if (dis[t.x][t.y] > dis[u.x][u.y] + 1) {
            next[cnt++] = t;
            dis[t.x][t.y] = dis[u.x][u.y] + 1;
            tway[t.x][t.y] = tway[u.x][u.y] + maps[t.x][t.y];
          } else if (dis[t.x][t.y] == dis[u.x][u.y] + 1) {
            string tmp = tway[u.x][u.y] + maps[t.x][t.y];
            if (tmp < tway[t.x][t.y]) {
              tway[t.x][t.y] = tmp;
            }
          }
        }
      }
    }
    if (!ok) break;
    for (i = 0; i < cnt; i++) {
      que.push(next[i]);
    }
  }
  if (dis[T.x][T.y] != 100000000) {
    tway[T.x][T.y].erase(tway[T.x][T.y].length() - 1, 1);
    if (res > dis[T.x][T.y]) {
      res = dis[T.x][T.y];
      way = tway[T.x][T.y];
    } else if (res == dis[T.x][T.y] && tway[T.x][T.y] < way) {
      way = tway[T.x][T.y];
    }
  }
}
void dfs(int st, int deep, string ste) {
  int i, j;
  if (deep == k) {
    j = ste.length();
    for (i = 0; i < j; i++) {
      if (rec.find(ste[i]) != -1) break;
    }
    if (i == j) return;
    bfs(ste);
    return;
  }
  for (j = st + 1; j < 26; j++) {
    dfs(j, deep + 1, ste + char(j + 'a'));
  }
}
int main() {
  int i, j;
  cin >> n >> m >> k;
  for (i = 0; i < n; i++) {
    scanf("%s", maps[i]);
    for (j = 0; j < m; j++) {
      if (maps[i][j] == 'S') {
        S = Node(i, j);
      }
      if (maps[i][j] == 'T') {
        T = Node(i, j);
      }
    }
  }
  rec = "";
  for (i = 0; i < 4; i++) {
    Node t = T;
    t.x += dir[i][0];
    t.y += dir[i][1];
    if (t.x >= 0 && t.x < n && t.y >= 0 && t.y < m) {
      rec += maps[t.x][t.y];
    }
  }
  dfs(-1, 0, "");
  if (res == 100000000 && rec.find('S') != -1) {
    puts("-1");
  } else {
    cout << way << endl;
  }
  return 0;
}
