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 double eps = 1e-6;
const double pi = acos(-1.0);
template <class Z>
inline Z SQ(Z &a) {
  return a * a;
}
template <class Z>
inline bool checkmax(Z &a, Z b) {
  if (a == -1 || a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class Z>
inline bool checkmin(Z &a, Z b) {
  if (a == -1 || a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class Z>
inline bool mymax(Z &a, Z b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class Z>
inline bool mymin(Z &a, Z b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int cas = 1;
void printcase() { printf("Case %d: ", cas++); }
struct node {
  int x, y, t;
  int sta, cnt;
  node() {}
  node(int __x, int __y, int __t, int __sta, int __cnt) {
    x = __x;
    y = __y;
    t = __t;
    sta = __sta;
    cnt = __cnt;
  }
};
map<int, int> dis[55][55];
queue<node> que;
int anstime, anssta;
int n, m, k;
int sx, sy, dx, dy;
int go[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
char str[55][55];
void init() {
  anstime = 0x7f7f7f7f;
  while (!que.empty()) que.pop();
  for (int i = 0; i < 55; i++) {
    for (int j = 0; j < 55; j++) {
      dis[i][j].clear();
    }
  }
}
void bfs() {
  que.push(node(sx, sy, 0, 0, 0));
  dis[sx][sy][0] = 0;
  while (!que.empty()) {
    node temp = que.front();
    que.pop();
    for (int i = 0; i < 4; i++) {
      int x = temp.x + go[i][0];
      int y = temp.y + go[i][1];
      int t = temp.t + 1;
      int cnt = temp.cnt;
      int sta = temp.sta;
      if (x < 0 || y < 0 || x >= n || y >= m) continue;
      if (x == dx && y == dy) {
        if (anstime > t) {
          dis[x][y][sta] = t;
          anstime = t;
          continue;
        } else if (anstime < t)
          return;
      }
      if (str[x][y] != 'S' && str[x][y] != 'T') {
        int num = str[x][y] - 'a';
        if ((sta & (1 << num)) == 0) {
          sta |= (1 << num);
          cnt++;
        }
      }
      if (cnt > k) continue;
      if (dis[x][y].find(sta) == dis[x][y].end()) {
        dis[x][y][sta] = t;
      } else if (dis[x][y][sta] > t) {
        dis[x][y][sta] = t;
      } else if (dis[x][y][sta] <= t)
        continue;
      que.push(node(x, y, t, sta, cnt));
    }
  }
}
char ans[10000];
struct nexnode {
  char ch;
  int idx;
  nexnode() {}
  nexnode(char __ch, int __idx) {
    ch = __ch;
    idx = __idx;
  }
  friend bool operator<(const nexnode &a, const nexnode &b) {
    return a.ch < b.ch;
  }
};
bool dfs(int si, int sj, int dep, int sta, int cnt) {
  if (dep > anstime) return false;
  if (dep + abs(si - dx) + abs(sj - dy) > anstime) return false;
  if (si == dx && sj == dy) {
    ans[dep - 1] = 0;
    puts(ans);
    return true;
  }
  int cc = 0;
  nexnode s[4];
  for (int i = 0; i < 4; i++) {
    int x = si + go[i][0];
    int y = sj + go[i][1];
    if (x < 0 || y < 0 || x >= n || y >= m) continue;
    if (str[x][y] == 'S') continue;
    s[cc++] = nexnode(str[x][y], i);
  }
  sort(s, s + cc);
  for (int i = 0; i < cc; i++) {
    int x = si + go[s[i].idx][0];
    int y = sj + go[s[i].idx][1];
    int num = str[x][y] - 'a';
    int nexsta = sta;
    int nexcnt = cnt;
    int t = dep + 1;
    if (str[x][y] != 'S' && str[x][y] != 'T') {
      if ((nexsta & (1 << num)) == 0) {
        nexsta |= (1 << num);
        nexcnt++;
      }
    }
    if (nexcnt > k) continue;
    if (dis[x][y].find(nexsta) == dis[x][y].end()) continue;
    if (dis[x][y][nexsta] == t) {
      ans[dep] = str[x][y];
      if (dfs(x, y, t, nexsta, nexcnt)) return true;
    } else
      continue;
  }
  return false;
}
int main() {
  while (~scanf("%d%d%d", &n, &m, &k)) {
    init();
    for (int i = 0; i < n; i++) {
      scanf("%s", str[i]);
      for (int j = 0; j < m; j++) {
        if (str[i][j] == 'S')
          sx = i, sy = j;
        else if (str[i][j] == 'T')
          dx = i, dy = j;
      }
    }
    bfs();
    if (anstime == 0x7f7f7f7f) {
      puts("-1");
    } else {
      if (!dfs(sx, sy, 0, 0, 0)) puts("");
    }
  }
}
