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 MN = 51;
const int inf = (1 << 30);
struct block {
  int x, y, mask;
  bool operator<(const block &nw) const {
    return (x < nw.x || (x == nw.x && y < nw.y) ||
            (x == nw.x && y == nw.y && mask < nw.mask));
  }
  static block make(int x, int y, int mask) {
    block nw;
    nw.x = x;
    nw.y = y;
    nw.mask = mask;
    return nw;
  }
};
struct block2 {
  int x, y;
  char ch;
  bool operator<(const block2 &nw) const { return (ch > nw.ch); }
  static block2 make(char ch, int x, int y) {
    block2 nw;
    nw.ch = ch;
    nw.x = x;
    nw.y = y;
    return nw;
  }
};
struct block3 {
  block b;
  int d, k;
  static block3 make(block b, int d, int k) {
    block3 nw;
    nw.b = b;
    nw.d = d;
    nw.k = k;
    return nw;
  }
};
struct block4 {
  int x, y, d;
  char ch;
  bool operator<(const block4 &nw) const {
    return (d > nw.d || (d == nw.d && ch > nw.ch));
  }
  static block4 make(int x, int y, int d, char ch) {
    block4 nw;
    nw.x = x;
    nw.y = y;
    nw.ch = ch;
    nw.d = d;
    return nw;
  }
};
map<block, int> mp;
char arr[MN][MN];
int n, m, k, ans = inf, ans_mask;
int px[] = {-1, 1, 0, 0};
int py[] = {0, 0, -1, 1};
int dp[MN][MN];
pair<int, int> st, fin, pr[MN][MN];
bool is_set(int mask, char ch) {
  int x = (1 << (ch - 'a'));
  return (mask & x);
}
int my_set(int mask, char ch) {
  int x = (1 << (ch - 'a'));
  mask |= x;
  return mask;
}
int cnt(int mask) {
  int sum = 0;
  for (int i = 0; i < 26; i++) {
    int x = (1 << i);
    if (x & mask) sum++;
  }
  return sum;
}
void solve(block curr, int curr_d, int curr_k) {
  queue<block3> q;
  q.push(block3::make(curr, curr_d, curr_k));
  while (!q.empty()) {
    block3 nw = q.front();
    q.pop();
    curr = nw.b;
    curr_d = nw.d;
    curr_k = nw.k;
    if (arr[curr.x][curr.y] == 'T') {
      ans = curr_d;
      ans_mask = curr.mask;
      break;
    }
    vector<block2> v;
    for (int i = 0; i < 4; i++) {
      int nx = curr.x + px[i];
      int ny = curr.y + py[i];
      if (nx >= 0 && ny >= 0 && nx < n && ny < m)
        v.push_back(block2::make(arr[nx][ny], nx, ny));
    }
    make_heap(v.begin(), v.end());
    while (!v.empty()) {
      int nx = v[0].x;
      int ny = v[0].y;
      pop_heap(v.begin(), v.end());
      v.pop_back();
      if (is_set(curr.mask, arr[nx][ny]) || arr[nx][ny] == 'T') {
        if (!mp.count(block::make(nx, ny, curr.mask)) &&
            (dp[nx][ny] == -1 || cnt(curr.mask) <= dp[nx][ny])) {
          q.push(
              block3::make(block::make(nx, ny, curr.mask), curr_d + 1, curr_k));
          mp[block::make(nx, ny, curr.mask)] = curr_d;
          dp[nx][ny] = cnt(curr.mask);
        }
      } else {
        int msk = my_set(curr.mask, arr[nx][ny]);
        if (curr_k - 1 >= 0 && !mp.count(block::make(nx, ny, msk)) &&
            (dp[nx][ny] == -1 || cnt(msk) <= dp[nx][ny])) {
          q.push(
              block3::make(block::make(nx, ny, msk), curr_d + 1, curr_k - 1));
          mp[block::make(nx, ny, msk)] = curr_d;
          dp[nx][ny] = cnt(msk);
        }
      }
    }
  }
}
string ans1, ans2;
void bfs(int x, int y, char q1) {
  vector<block4> pq;
  queue<block> q;
  q.push(block::make(x, y, 0));
  fill(&dp[0][0], &dp[0][0] + MN * MN, -1);
  fill(&pr[0][0], &pr[0][0] + MN * MN, make_pair(-1, -1));
  dp[x][y] = 0;
  pq.push_back(block4::make(x, y, 0, 'a'));
  while (!pq.empty()) {
    block4 nw = pq[0];
    pop_heap(pq.begin(), pq.end());
    pq.pop_back();
    x = nw.x;
    y = nw.y;
    if (arr[x][y] == q1) break;
    vector<block2> v;
    for (int i = 0; i < 4; i++) {
      int nx = nw.x + px[i];
      int ny = nw.y + py[i];
      if (nx >= 0 && ny >= 0 && nx < n && ny < m)
        v.push_back(block2::make(arr[nx][ny], nx, ny));
    }
    make_heap(v.begin(), v.end());
    while (!v.empty()) {
      int nx = v[0].x;
      int ny = v[0].y;
      char ch = arr[nx][ny];
      pop_heap(v.begin(), v.end());
      v.pop_back();
      if ((arr[nx][ny] == q1 || is_set(ans_mask, ch)) && dp[nx][ny] == -1) {
        dp[nx][ny] = dp[x][y] + 1;
        pr[nx][ny] = make_pair(x, y);
        pq.push_back(block4::make(nx, ny, dp[x][y] + 1, ch));
        push_heap(pq.begin(), pq.end());
      }
    }
  }
}
bool flag = false;
string ans3;
vector<string> vs;
char str[MN * MN];
int dl;
void dfs(int x, int y, int dist, int deep) {
  if (vs.size() > 300) return;
  if (deep > dist) return;
  if (arr[x][y] == 'T') {
    string tmp;
    for (int i = 0; i < dl; i++)
      if (str[i] != 'T' && str[i] != 'S') tmp.push_back(str[i]);
    vs.push_back(tmp);
    return;
  }
  vector<block2> v;
  for (int i = 0; i < 4; i++) {
    int nx = x + px[i];
    int ny = y + py[i];
    if (nx >= 0 && ny >= 0 && nx < n && ny < m)
      v.push_back(block2::make(arr[nx][ny], nx, ny));
  }
  make_heap(v.begin(), v.end());
  while (!v.empty()) {
    int nx = v[0].x;
    int ny = v[0].y;
    char ch = arr[nx][ny];
    pop_heap(v.begin(), v.end());
    v.pop_back();
    if ((arr[nx][ny] == 'T' || is_set(ans_mask, ch)) &&
        dp[nx][ny] == dp[x][y] + 1) {
      if (arr[x][y] != 'S') str[dl++] = arr[x][y];
      dfs(nx, ny, dist, deep + 1);
      if (arr[x][y] != 'S') dl--;
    }
  }
}
void get_path(int x, int y, string &s, char q1, char q2) {
  if (arr[x][y] != q1) {
    get_path(pr[x][y].first, pr[x][y].second, s, q1, q2);
    if (arr[x][y] != q2) s.push_back(arr[x][y]);
  }
}
int main() {
  cin >> n >> m >> k;
  fill(&pr[0][0], &pr[0][0] + MN * MN, make_pair(-1, -1));
  fill(&dp[0][0], &dp[0][0] + MN * MN, -1);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
      if (arr[i][j] == 'S')
        st = make_pair(i, j);
      else if (arr[i][j] == 'T')
        fin = make_pair(i, j);
    }
  solve(block::make(st.first, st.second, 0), 0, k);
  if (ans == inf)
    cout << -1;
  else {
    bfs(st.first, st.second, 'T');
    dfs(st.first, st.second, ans, 0);
    sort(vs.begin(), vs.end());
    cout << vs[0];
  }
  return 0;
}
