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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 51;
int n, m, K;
int S[2], T[2];
char smap[N][N];
bool bk[256];
int C[256];
char M[4];
char lala[4];
void dfs(int step) {
  if (step > K) {
    bool lala(0);
    for (int i = 0; i < 4; ++i) lala |= bk[M[i]];
    if (!lala) return;
    void update();
    update();
    return;
  }
  for (int i = 'z'; i >= 'a'; --i)
    if (!bk[i]) {
      bk[i] = 1;
      C[i] = K - step;
      lala[K - step] = i;
      dfs(step + 1);
      bk[i] = 0;
    }
}
char lst[N << 5];
int ans = 0x3f3f3f3f;
int dis[N][N];
struct STT {
  unsigned long long a[8];
  STT() { memset(a, 0, sizeof(a)); }
  friend STT operator*(int l, STT ss) {
    STT s = ss;
    for (int i = 0; i < 7; ++i) s.a[i] <<= 2, s.a[i] |= s.a[i + 1] >> 62;
    s.a[7] <<= 2, s.a[7] |= l;
    return s;
  }
  friend bool operator<(STT s1, STT s2) {
    for (int i = 0; i < 8; ++i)
      if (s1.a[i] < s2.a[i])
        return 1;
      else if (s1.a[i] > s2.a[i])
        return 0;
    return 0;
  }
  friend bool operator>(STT s1, STT s2) {
    for (int i = 0; i < 8; ++i)
      if (s1.a[i] > s2.a[i])
        return 1;
      else if (s1.a[i] < s2.a[i])
        return 0;
    return 0;
  }
  friend bool operator==(STT s1, STT s2) { return !(s1 > s2 || s1 < s2); }
} stt[N][N];
int XX[4] = {0, 0, 1, -1};
int YY[4] = {1, -1, 0, 0};
char tmplst[N << 5];
void update() {
  memset(dis, -1, sizeof(dis));
  memset(stt, 0, sizeof(stt));
  dis[S[0]][S[1]] = 0;
  queue<int> q[2];
  q[0].push(S[0]);
  q[1].push(S[1]);
  while (!q[0].empty()) {
    int nx = q[0].front();
    int ny = q[1].front();
    q[0].pop(), q[1].pop();
    for (int i = 0; i < 4; ++i) {
      int x = nx + XX[i];
      int y = ny + YY[i];
      if (x < 1 || y < 1 || x > n || y > m ||
          (!bk[smap[x][y]] && smap[x][y] != 'T'))
        continue;
      if (smap[x][y] == 'T') {
        if (dis[x][y] == -1)
          stt[x][y] = stt[nx][ny], dis[x][y] = dis[nx][ny];
        else if (dis[x][y] > dis[nx][ny]) {
          stt[x][y] = stt[nx][ny];
          dis[x][y] = dis[nx][ny];
        } else if (dis[x][y] == dis[nx][ny] && stt[x][y] > stt[nx][ny])
          stt[x][y] = stt[nx][ny];
        continue;
      }
      if (dis[x][y] == -1) {
        stt[x][y] = C[smap[x][y]] * stt[nx][ny];
        dis[x][y] = dis[nx][ny] + 1;
        q[0].push(x);
        q[1].push(y);
      } else if (dis[x][y] > dis[nx][ny] + 1) {
        stt[x][y] = C[smap[x][y]] * stt[nx][ny];
        dis[x][y] = dis[nx][ny] + 1;
      } else if (dis[x][y] == dis[nx][ny] + 1 &&
                 stt[x][y] > C[smap[x][y]] * stt[nx][ny]) {
        stt[x][y] = C[smap[x][y]] * stt[nx][ny];
      }
    }
  }
  if (dis[T[0]][T[1]] == -1) return;
  if (dis[T[0]][T[1]] <= ans) {
    int la(STT, int);
    if (ans > dis[T[0]][T[1]]) {
      ans = dis[T[0]][T[1]];
      for (int i = ans; i; --i) lst[i] = lala[la(stt[T[0]][T[1]], ans - i)];
    } else {
      ans = dis[T[0]][T[1]];
      for (int i = ans; i; --i) tmplst[i] = lala[la(stt[T[0]][T[1]], ans - i)];
      if (strcmp(tmplst + 1, lst + 1) < 0)
        for (int i = 1; i <= ans; ++i) lst[i] = tmplst[i];
    }
  }
}
int la(STT s, int k) {
  int i = 7 - k / 32;
  k %= 32;
  return 3 & (s.a[i] >> (2 * k));
}
void pt() {
  if (ans == 0x3f3f3f3f) {
    puts("-1");
    return;
  }
  for (int i = 1; i <= ans; ++i) putchar(lst[i]);
  putchar('\n');
}
int main() {
  n = read(), m = read(), K = read();
  for (int i = 1; i <= n; ++i) scanf("%s", smap[i] + 1);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (smap[i][j] == 'S')
        S[0] = i, S[1] = j;
      else if (smap[i][j] == 'T')
        T[0] = i, T[1] = j;
  for (int i = 0, x, y; i < 4; ++i) {
    x = S[0] + XX[i], y = S[1] + YY[i];
    if (x < 1 || y < 1 || x > n || y > m) continue;
    M[i] = smap[x][y];
  }
  dfs(1);
  pt();
}
