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 map[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
struct node {
  int x, y;
};
char a[60][60];
node s, t, zhan[3000], zhan2[3000];
int pan[60][60], pan2[60][60], dui[30], pdui[10], id, len[60][60];
int n, m, ss[3000], ans[3000];
bool use[30], check;
void getans(int x) {
  int j, x1, y1, x2, y2, s1, t1, v1, i, p, ge, ge2, l, kk;
  kk = pdui[2];
  for (j = 0; j < 26; j++) use[j] = false;
  for (j = 1; j <= x; j++) use[dui[pdui[j]]] = true;
  id++;
  zhan[1] = t;
  pan[t.x][t.y] = id;
  len[t.x][t.y] = 0;
  s1 = t1 = v1 = 1;
  while (s1 <= t1) {
    for (j = s1; j <= t1; j++) {
      x1 = zhan[j].x;
      y1 = zhan[j].y;
      for (i = 0; i < 4; i++) {
        x2 = x1 + map[i][0];
        y2 = y1 + map[i][1];
        if (x2 == s.x && y2 == s.y) {
          len[x2][y2] = len[x1][y1] + 1;
          pan[x2][y2] = id;
        } else if (x2 > 0 && x2 <= n && y2 > 0 && y2 <= m &&
                   use[a[x2][y2 - 1] - 'a'] && pan[x2][y2] < id) {
          v1++;
          zhan[v1].x = x2;
          zhan[v1].y = y2;
          pan[x2][y2] = id;
          len[x2][y2] = len[x1][y1] + 1;
        }
      }
    }
    s1 = t1 + 1;
    t1 = v1;
  }
  ss[0] = 0;
  if (pan[s.x][s.y] == id) {
    check = true;
    zhan[1] = s;
    ge = 1;
    while (1) {
      ge2 = 0;
      l = 26;
      for (j = 1; j <= ge; j++) {
        x1 = zhan[j].x;
        y1 = zhan[j].y;
        for (i = 0; i < 4; i++) {
          x2 = x1 + map[i][0];
          y2 = y1 + map[i][1];
          if (x2 == t.x && y2 == t.y) {
            if (ss[0] < ans[0]) {
              ans[0] = ss[0];
              for (p = 1; p <= ss[0]; p++) ans[p] = ss[p];
            } else if (ss[0] == ans[0]) {
              int xi = 0;
              for (p = 1; p <= ss[0]; p++) {
                if (ss[p] < ans[p]) {
                  xi = -1;
                  break;
                } else if (ss[p] > ans[p]) {
                  xi = 1;
                  break;
                }
              }
              if (xi < 0) {
                for (p = 1; p <= ss[0]; p++) ans[p] = ss[p];
              }
            }
            return;
          }
          if (x2 > 0 && x2 <= n && y2 > 0 && y2 <= m && pan[x2][y2] == id &&
              len[x2][y2] == len[x1][y1] - 1)
            if (pan2[x2][y2] < id) {
              ge2++;
              zhan2[ge2].x = x2;
              zhan2[ge2].y = y2;
              pan2[x2][y2] = id;
              if (a[x2][y2 - 1] - 'a' < l) l = a[x2][y2 - 1] - 'a';
            }
        }
      }
      ss[0]++;
      ss[ss[0]] = l;
      ge = 0;
      for (j = 1; j <= ge2; j++)
        if (a[zhan2[j].x][zhan2[j].y - 1] - 'a' == l) {
          ge++;
          zhan[ge] = zhan2[j];
        }
    }
  }
}
void go(int x) {
  int j, t1;
  for (j = 1; j <= x; j++) pdui[j] = j;
  pdui[0] = 0;
  while (pdui[0] == 0) {
    getans(x);
    t1 = x;
    while (t1 > 0 && pdui[t1] == dui[0] - x + t1) t1--;
    pdui[t1]++;
    if (t1 != 0) {
      for (j = t1 + 1; j <= x; j++) pdui[j] = pdui[j - 1] + 1;
    }
  }
}
int main() {
  int k, j, i, len;
  scanf("%d%d%d", &n, &m, &k);
  for (j = 0; j < 26; j++) use[j] = false;
  dui[0] = 0;
  for (j = 1; j <= n; j++) {
    scanf("%s", a[j]);
    len = strlen(a[j]);
    for (i = 0; i < len; i++) {
      if (a[j][i] != 'S' && a[j][i] != 'T')
        if (!use[a[j][i] - 'a']) {
          dui[0]++;
          dui[dui[0]] = a[j][i] - 'a';
          use[a[j][i] - 'a'] = true;
        }
      if (a[j][i] == 'S') {
        s.x = j;
        s.y = i + 1;
      }
      if (a[j][i] == 'T') {
        t.x = j;
        t.y = i + 1;
      }
    }
  }
  k = min(k, dui[0]);
  if (k != 0) {
    check = false;
    ans[0] = 3000;
    for (j = 1; j <= n; j++)
      for (i = 1; i <= m; i++) pan[j][i] = pan2[j][i] = 0;
    id = 0;
    for (j = 1; j <= k; j++) {
      go(j);
    }
    if (!check)
      printf("-1\n");
    else if (ans[0] != 0) {
      for (j = 1; j <= ans[0]; j++) printf("%c", (char)(ans[j] + 'a'));
    }
  }
  return 0;
}
