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;
set<pair<int, int> > ss;
vector<int> adj[3000], v, vs, ve;
int a[3000];
int p[3000];
bool vis[3000], flag, f;
int n, m, k, kk;
vector<int> vc(10);
int start, ed;
string ans[10], s;
int tot;
bool cmp(int i, int j) { return a[i] < a[j]; }
void build() {
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      int po = n * (i - 1) + j;
      if (ss.count(pair<int, int>(i, j - 1)))
        adj[po].push_back(n * (i - 1) + j - 1);
      if (ss.count(pair<int, int>(i, j + 1)))
        adj[po].push_back(n * (i - 1) + j + 1);
      if (ss.count(pair<int, int>(i - 1, j)))
        adj[po].push_back(n * (i - 2) + j);
      if (ss.count(pair<int, int>(i + 1, j))) adj[po].push_back(n * i + j);
      sort(adj[po].begin(), adj[po].end(), cmp);
    }
  }
}
bool chk(int v) {
  int po = lower_bound(vc.begin(), vc.begin() + kk, a[v]) - vc.begin();
  if (vc[po] == a[v])
    return true;
  else
    return false;
}
void BFS(int st, int en) {
  for (int i = 1; i <= n * m; i++) {
    vis[i] = 0;
  }
  vis[st] = 1;
  if (st == en) {
    flag = 1;
    return;
  }
  queue<int> Q;
  vector<int> sv;
  Q.push(st);
  while (!Q.empty()) {
    sv.clear();
    while (!Q.empty()) {
      sv.push_back(Q.front());
      Q.pop();
    }
    sort(sv.begin(), sv.end(), cmp);
    for (int i = 0; i < sv.size(); i++) Q.push(sv[i]);
    int u = Q.front();
    Q.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (v == 1) continue;
      if (v == n * m) continue;
      if (v == en) {
        flag = 1;
        p[en] = u;
        break;
      }
      if (vis[v] == 0 && chk(v)) {
        vis[v] = 1;
        p[v] = u;
        Q.push(v);
      }
    }
    vis[u] = 1;
    if (flag) {
      break;
    }
  }
}
void track(int st, int len) {
  if (len == kk) {
    flag = 0;
    BFS(start, ed);
    if (flag == 1) {
      f = 1;
      int u = ed;
      ans[tot] += a[u];
      while (u != start) {
        u = p[u];
        ans[tot] += a[u];
      }
      tot++;
    }
    return;
  }
  for (int i = st; i < v.size(); i++) {
    vc[len] = v[i];
    track(i, len + 1);
    if (flag) return;
  }
}
int main() {
  char c;
  scanf("%d%d%d", &m, &n, &k);
  if (m * n == 2) {
    puts("");
    return 1;
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      scanf("%c", &c);
      if (c == '\n') {
        j--;
        continue;
      }
      a[n * (i - 1) + j] = c;
      ss.insert(pair<int, int>(i, j));
    }
  }
  build();
  vector<int> st, en;
  vs.push_back(a[2]), st.push_back(2);
  if (n + 1 < n * m) vs.push_back(a[n + 1]), st.push_back(n + 1);
  ve.push_back(a[n * m - 1]), en.push_back(n * m - 1);
  if (n * (m - 1) > 1) ve.push_back(a[n * (m - 1)]), en.push_back(n * (m - 1));
  kk = k;
  for (int i = 0; i < st.size(); i++) {
    for (int j = 0; j < en.size(); j++) {
      vc.clear();
      k = kk;
      if (vs[i] == ve[j])
        k--, vc.push_back(vs[i]);
      else if (k < 2)
        continue;
      else if (k >= 2) {
        vc.push_back(vs[i]), vc.push_back(ve[j]), k -= 2;
      }
      v.clear();
      for (int l = 'a'; l <= 'z'; l++) {
        if (l == vs[i] || l == ve[j]) continue;
        v.push_back(l);
      }
      start = st[i];
      ed = en[j];
      track(0, vc.size());
    }
  }
  if (f) {
    s = ans[0];
    reverse(s.begin(), s.end());
    for (int i = 1; i < tot; i++) {
      reverse(ans[i].begin(), ans[i].end());
      if (s > ans[i]) s = ans[i];
    }
    cout << s;
  } else
    puts("-1");
}
