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 INF = 0x3f3f3f3f;
struct Pos {
  int x : 8;
  int y : 8;
  Pos() {}
  Pos(int x, int y) : x(x), y(y) {}
};
bool operator!=(const Pos &A, const Pos &B) { return A.x != B.x || A.y != B.y; }
char field[55][55];
bool can[26], debug;
int n, m, k;
int sx, sy, tx, ty;
string ans;
vector<char> vt, cvt;
vector<Pos> pre[55][55];
bool vi[55][55];
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
bool out(int x, int y) { return x >= n || x < 0 || y >= m || y < 0; }
void getPath(int x, int y, string &path, int now) {
  vi[x][y] = true;
  for (vector<Pos>::iterator it = pre[x][y].begin(); it != pre[x][y].end();
       it++) {
    if (it->x == tx && it->y == ty) {
      return;
    }
    if (field[it->x][it->y] > path[now] || vi[it->x][it->y]) {
      continue;
    }
    path[now] = field[it->x][it->y];
    getPath(it->x, it->y, path, now + 1);
  }
}
void cal() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      pre[i][j].clear();
    }
  }
  int dist[55][55];
  fill(dist[0], dist[n], INF);
  dist[tx][ty] = 0;
  queue<Pos> Q;
  Q.push(Pos(tx, ty));
  while (!Q.empty()) {
    Pos now = Q.front();
    Q.pop();
    for (int i = 0; i < 4; i++) {
      Pos next = now;
      next.x += dx[i];
      next.y += dy[i];
      if (out(next.x, next.y) ||
          (field[next.x][next.y] != 'S' && !can[field[next.x][next.y] - 'a'])) {
        continue;
      }
      vector<Pos> &pren = pre[next.x][next.y];
      if (dist[next.x][next.y] > dist[now.x][now.y] + 1) {
        dist[next.x][next.y] = dist[now.x][now.y] + 1;
        pren.clear();
        pren.push_back(now);
        Q.push(next);
      } else if (dist[next.x][next.y] == dist[now.x][now.y] + 1) {
        Pos p = pren.front();
        if (field[now.x][now.y] < field[p.x][p.y]) {
          pren.clear();
          pren.push_back(now);
        } else if (field[now.x][now.y] == field[p.x][p.y]) {
          pren.push_back(now);
        }
      }
    }
  }
  if (dist[sx][sy] == INF || ans != "~" && dist[sx][sy] - 1 > ans.length()) {
    return;
  }
  memset(vi, 0, sizeof vi);
  string path = string(dist[sx][sy] - 1, 'z');
  getPath(sx, sy, path, 0);
  if (ans == "~" || dist[sx][sy] - 1 < ans.length()) {
    ans = path;
    return;
  }
  if (ans > path) {
    ans = path;
    return;
  }
}
void dfs(int pos, int have) {
  if (have == k || have == vt.size()) {
    fill(can, can + 26, false);
    for (vector<char>::iterator it = cvt.begin(); it != cvt.end(); it++) {
      can[*it - 'a'] = true;
    }
    cal();
    return;
  }
  for (int i = pos; i < vt.size(); i++) {
    cvt.push_back(vt[i]);
    dfs(i + 1, have + 1);
    cvt.pop_back();
  }
}
int main() {
  for (; ~scanf("%d%d%d", &n, &m, &k);) {
    vt.clear();
    for (int i = 0; i < n; i++) {
      scanf("%s", field[i]);
      for (int j = 0; j < m; j++) {
        if (field[i][j] == 'S') {
          sx = i;
          sy = j;
        } else if (field[i][j] == 'T') {
          tx = i;
          ty = j;
        } else {
          vt.push_back(field[i][j]);
        }
      }
    }
    sort(vt.begin(), vt.end());
    vt.resize(unique(vt.begin(), vt.end()) - vt.begin());
    ans = "~";
    cvt.clear();
    dfs(0, 0);
    if (ans == "~") {
      cout << "-1\n";
    } else {
      cout << ans << endl;
    }
  }
  return 0;
}
