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 long long LINF = 1e17 + 7;
const unsigned long long BASE = 33;
const int N = 55;
const int INF = 1e9 + 7;
const int MOD = 2e9 + 7;
const double Pi = acos(-1.);
const double EPS = 1e-8;
long long kpow(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, a = a * a)
    if (b & 1) ret = ret * a;
  return ret;
}
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int n, m, k;
pair<int, int> S, T;
char ch[5], Map[N][N];
int a[N][N], dis[2][N][N];
int que[N * N], nxt[N * N];
int top, len;
char tmp[N * N], ans[N * N];
bool check() {
  if (len == 0 || top < len) return true;
  if (top > len) return false;
  for (int i = 0; i < (len); ++i)
    if (tmp[i] != ans[i]) return tmp[i] < ans[i];
  return false;
}
void upd() {
  top = 0;
  int qh, qt, nh, nt;
  qh = qt = 0, nh = nt = 0;
  que[qt++] = S.first << 6 | S.second;
  int st = dis[0][T.first][T.second];
  for (int d = 0; d < (dis[0][T.first][T.second]); ++d) {
    int mx = k;
    nh = nt = 0;
    while (qh < qt) {
      int sx, sy;
      sx = que[qh] >> 6;
      sy = que[qh] ^ (sx << 6);
      ++qh;
      for (int i = 0; i < (4); ++i) {
        int x = sx + dx[i], y = sy + dy[i];
        if (!(0 < x && x <= n && 0 < y && y <= m)) continue;
        if (dis[0][x][y] != d + 1 || dis[1][x][y] + d + 1 != st) continue;
        mx = min(mx, a[x][y]);
        nxt[nt++] = x << 6 | y;
      }
    }
    tmp[top++] = ch[mx];
    qh = qt = 0;
    while (nh < nt) {
      int sx, sy;
      sx = nxt[nh] >> 6;
      sy = nxt[nh] ^ (sx << 6);
      ++nh;
      if (a[sx][sy] == mx) que[qt++] = sx << 6 | sy;
    }
  }
  if (check()) {
    len = top;
    for (int i = 0; i < (top); ++i) ans[i] = tmp[i];
  }
}
void solve(pair<int, int> S, pair<int, int> T, int dis[][N]) {
  int qh = 0, qt = 0;
  for (int i = 1; i < (n + 1); ++i)
    for (int j = 1; j < (m + 1); ++j) dis[i][j] = -1;
  dis[S.first][S.second] = 0, que[qt++] = S.first << 6 | S.second;
  while (qh < qt) {
    int sx, sy;
    sx = que[qh] >> 6;
    sy = que[qh] ^ (sx << 6);
    ++qh;
    if (sx == T.first && sy == T.second) continue;
    for (int i = 0; i < (4); ++i) {
      int x = sx + dx[i], y = sy + dy[i];
      if (!(0 < x && x <= n && 0 < y && y <= m)) continue;
      if (a[x][y] == -1 || ~dis[x][y]) continue;
      dis[x][y] = dis[sx][sy] + 1;
      que[qt++] = x << 6 | y;
    }
  }
}
void dfs(int t, int x) {
  if (x >= k) {
    for (int i = 1; i < (n + 1); ++i)
      for (int j = 1; j < (m + 1); ++j)
        if (Map[i][j] == 'S' || Map[i][j] == 'T') {
          a[i][j] = k;
        } else {
          a[i][j] = -1;
          for (int l = 0; l < (k); ++l)
            if (ch[l] == Map[i][j]) {
              a[i][j] = l;
              break;
            }
        }
    solve(S, T, dis[0]), solve(T, S, dis[1]);
    if (~dis[0][T.first][T.second]) upd();
    return;
  }
  if (t >= 26) return;
  ch[x] = 'a' + t, dfs(t + 1, x + 1);
  dfs(t + 1, x);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i < (n + 1); ++i) scanf(" %s", Map[i] + 1);
  for (int i = 1; i < (n + 1); ++i)
    for (int j = 1; j < (m + 1); ++j)
      if (Map[i][j] == 'S') {
        S = make_pair(i, j);
      } else if (Map[i][j] == 'T') {
        T = make_pair(i, j);
      }
  ch[k] = '#';
  dfs(0, 0);
  if (!len) {
    puts("-1");
  } else {
    for (int i = 0; i < (len - 1); ++i) putchar(ans[i]);
  }
  return 0;
}
