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>
char ch[60][60], path[60 * 60], answer[60 * 60];
int n, m, K, alpha[60], dir[60][60][5], child[60][60];
int sx, sy, ex, ey, cnt = 2147483647;
int tlimit;
bool check[60][60], flag = false;
int dx[5] = {0, 1, 0, -1, 0}, dy[5] = {0, 0, 1, 0, -1};
inline int abs(int t) { return (t > 0) ? t : -t; }
void dfs(int x, int y, int dif, int dep) {
  int i, x1, y1, t;
  tlimit++;
  if (dep + abs(x - ex) + abs(y - ey) >= cnt) return;
  if (x == ex && y == ey) {
    flag = true;
    if (dep < cnt) {
      cnt = dep;
      strcpy(&answer[1], &path[1]);
    } else if (dep == cnt) {
      if (strcmp(&answer[1], &path[1]) > 0) strcpy(&answer[1], &path[1]);
    }
  } else {
    if (dep != 0) {
      path[dep] = ch[x][y];
      t = ch[x][y] - 'a' + 1;
      alpha[t]++;
    }
    check[x][y] = true;
    for (i = 1; i <= child[x][y]; i++) {
      x1 = x + dx[dir[x][y][i]], y1 = y + dy[dir[x][y][i]];
      if (check[x1][y1]) continue;
      t = ch[x1][y1] - 'a' + 1;
      if (alpha[t] != 0 || dep == 0) {
        dfs(x1, y1, dif, dep + 1);
        if (tlimit >= 10000000) return;
      } else {
        if (dif == K) continue;
        dfs(x1, y1, dif + 1, dep + 1);
        if (tlimit >= 10000000) return;
      }
    }
    check[x][y] = false;
    if (dep != 0) path[dep] = 0;
    t = ch[x][y] - 'a' + 1;
    alpha[t]--;
  }
}
struct data {
  char c;
  int idx;
};
inline bool sortf(const data &a, const data &b) {
  if (a.c < b.c) return true;
  return false;
}
int main(void) {
  int i, j, x, y, k, ccnt;
  data cc[5];
  scanf("%d %d %d", &n, &m, &K);
  for (i = 1; i <= n; i++) {
    scanf("%s", &ch[i][1]);
    for (j = 1; j <= m; j++) {
      if (ch[i][j] == 'S') sx = i, sy = j;
      if (ch[i][j] == 'T') ex = i, ey = j;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      ccnt = 0;
      for (k = 1; k <= 4; k++) {
        x = i + dx[k], y = j + dy[k];
        if (x > n || x < 1 || y > m || y < 1) continue;
        if (x == sx && y == sy) continue;
        if (x == ex && y == ey) {
          child[i][j]++;
          dir[i][j][1] = k;
          break;
        }
      }
      for (k = 1; k <= 4; k++) {
        x = i + dx[k], y = j + dy[k];
        if (x > n || x < 1 || y > m || y < 1) continue;
        if (x == sx && y == sy) continue;
        if (x == ex && y == ey) continue;
        if (ch[x][y] == ch[i][j]) {
          child[i][j]++;
          dir[i][j][child[i][j]] = k;
        } else {
          cc[++ccnt].c = ch[x][y];
          cc[ccnt].idx = k;
        }
      }
      std::sort(cc + 1, cc + ccnt + 1, sortf);
      for (k = 1; k <= ccnt; k++) {
        child[i][j]++;
        dir[i][j][child[i][j]] = cc[k].idx;
      }
    }
  }
  dfs(sx, sy, 0, 0);
  if (!flag)
    printf("-1\n");
  else
    printf("%s\n", &answer[1]);
  return false;
}
