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;
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int n, m, k, na, nb, nc, S, T;
const int md = 1e6 + 3;
char c[6666666];
int head[6666666], nxt[6666666], to[6666666], dist[6666666], sx, sy, tx, ty,
    cnt, cur, mto;
int headr[6666666], nxtr[6666666], tor[6666666], cntr;
char cm;
bool In(int x, int y) { return (x >= 0 && x < n && y >= 0 && y < m); }
struct hashtable {
  int head[md + 5], nxt[6666666], val[6666666], cnt;
  void add(int x, char ch) {
    int g = x % md;
    ++cnt;
    val[cnt] = x;
    c[cnt] = ch;
    nxt[cnt] = head[g];
    head[g] = cnt;
  }
  int find(int x, char c) {
    int g = x % md;
    for (int i = head[g]; i; i = nxt[i]) {
      if (val[i] == x) return i;
    }
    add(x, c);
    return cnt;
  }
} ht;
int get(int x, int y, int a, int b, int c) {
  return (x * 2000000 + y * 30000 + a * 1000 + b * 30 + c);
}
char s[55][55];
void addedge(int t, int s) {
  ++cnt;
  to[cnt] = t;
  nxt[cnt] = head[s];
  head[s] = cnt;
  ++cntr;
  tor[cnt] = s;
  nxtr[cntr] = headr[t];
  headr[t] = cntr;
}
void bfs() {
  memset(dist, -1, sizeof(dist));
  dist[T] = 0;
  queue<int> q;
  q.push(T);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = nxt[i]) {
      if (!~dist[to[i]]) {
        dist[to[i]] = dist[x] + 1;
        q.push(to[i]);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", s[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int a = 0; a <= 26; a++) {
        if (s[i][j] == 'S' && a) continue;
        if (s[i][j] == a + 'a' - 1) continue;
        for (int b = 0; b <= 26; b++) {
          if (a < b) continue;
          if (a && a == b) continue;
          if (s[i][j] == 'S' && b) continue;
          if (s[i][j] == b + 'a' - 1) continue;
          for (int c = 0; c <= 26; c++) {
            if (b < c) continue;
            if (b && b == c) continue;
            if (s[i][j] == 'S' && c) continue;
            if (s[i][j] == c + 'a' - 1) continue;
            int id = ht.find(get(i, j, a, b, c), s[i][j]);
            for (int p = 0; p < 4; p++) {
              int xx = i + dx[p], yy = j + dy[p];
              if (!In(xx, yy)) continue;
              if (s[xx][yy] == s[i][j]) {
                addedge(id, ht.find(get(xx, yy, a, b, c), s[xx][yy]));
              } else {
                int tt = s[xx][yy] - 'a' + 1, pp = s[i][j] - 'a' + 1;
                if (pp < 0) {
                  na = nb = nc = 0;
                } else if (tt < 0) {
                  na = a;
                  nb = b;
                  nc = c;
                } else if (a == tt) {
                  if (pp == b || pp == c) {
                    na = b;
                    nb = c;
                    nc = 0;
                  } else {
                    if (pp > b) {
                      na = pp;
                      nb = b;
                      nc = c;
                    } else if (pp > c) {
                      na = b;
                      nb = pp;
                      nc = c;
                    } else {
                      na = b;
                      nb = c;
                      nc = pp;
                    }
                  }
                } else if (b == tt) {
                  if (pp == a || pp == c) {
                    na = a;
                    nb = c;
                    nc = 0;
                  } else {
                    if (pp > a) {
                      na = pp;
                      nb = a;
                      nc = c;
                    } else if (pp > c) {
                      na = a;
                      nb = pp;
                      nc = c;
                    } else {
                      na = a;
                      nb = c;
                      nc = pp;
                    }
                  }
                } else if (c == tt) {
                  if (pp == a || pp == b) {
                    na = a;
                    nb = b;
                    nc = 0;
                  } else {
                    if (pp > a) {
                      na = pp;
                      nb = a;
                      nc = b;
                    } else if (pp > b) {
                      na = a;
                      nb = pp;
                      nc = b;
                    } else {
                      na = a;
                      nb = b;
                      nc = pp;
                    }
                  }
                } else {
                  if (pp == a || pp == b || pp == c) {
                    na = a;
                    nb = b;
                    nc = c;
                  } else {
                    if (!a) {
                      if (k <= 1) continue;
                      na = pp;
                    } else if (!b) {
                      if (k <= 2) continue;
                      if (pp > a) {
                        na = pp;
                        nb = a;
                        nc = 0;
                      } else {
                        na = a;
                        nb = pp;
                        nc = 0;
                      }
                    } else if (!c) {
                      if (k <= 3) continue;
                      if (pp > a) {
                        na = pp;
                        nb = a;
                        nc = b;
                      } else if (pp > b) {
                        na = a;
                        nb = pp;
                        nc = b;
                      } else {
                        na = a;
                        nb = b;
                        nc = pp;
                      }
                    } else
                      continue;
                  }
                }
                addedge(id, ht.find(get(xx, yy, na, nb, nc), s[xx][yy]));
              }
            }
          }
        }
      }
      if (s[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (s[i][j] == 'T') {
        tx = i;
        ty = j;
      }
    }
  }
  S = ht.find(get(sx, sy, 0, 0, 0), s[sx][sy]);
  T = ht.cnt + 1;
  for (int a = 0; a <= 26; a++) {
    for (int b = 0; b <= 26; b++) {
      if (a < b) continue;
      if (a && a == b) continue;
      for (int c = 0; c <= 26; c++) {
        if (b < c) continue;
        if (b && b == c) continue;
        addedge(ht.find(get(tx, ty, a, b, c), s[tx][ty]), T);
      }
    }
  }
  bfs();
  if (!~dist[S]) {
    printf("-1\n");
    return 0;
  }
  int cur = S;
  while (cur != T) {
    cm = 'z' + 1;
    for (int i = headr[cur]; i; i = nxtr[i]) {
      if (dist[tor[i]] + 1 != dist[cur]) continue;
      if (c[tor[i]] < cm) {
        cm = c[tor[i]];
        mto = tor[i];
      }
    }
    if (cm != 'T') printf("%c", cm);
    cur = mto;
  }
  printf("\n");
  return 0;
}
