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 MAXN = 1e2 + 5;
const double EPS = 1e-9;
const long long INF = 1000000000 + 5;
const int MAXN2 = 2e5 + 5;
bool generate(vector<int> &a, int n) {
  int k = a.size();
  for (int i = k - 1; i >= 0; i--) {
    if (a[i] < n - k + i + 1) {
      a[i]++;
      for (int j = i + 1; j < k; j++) {
        a[j] = a[j - 1] + 1;
      }
      return true;
    }
  }
  return false;
}
vector<int> a;
char s[MAXN][MAXN];
bool visited[MAXN][MAXN];
int p[MAXN][MAXN][2];
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
int d[MAXN][MAXN];
int n, m, k;
string ans = "";
bool getS(const pair<int, int> &a, const pair<int, int> &b) {
  int x1 = a.first, y1 = a.second;
  int x2 = b.first, y2 = b.second;
  if (s[x1][y1] < s[x2][y2])
    return true;
  else
    return false;
}
void dfs(int x2, int y2, int mask) {
  p[x2][y2][0] = -1;
  p[x2][y2][1] = -1;
  queue<int> qx, qy;
  visited[x2][y2] = 1;
  qx.push(x2);
  qy.push(y2);
  d[x2][y2] = 0;
  while (!qx.empty()) {
    int x = qx.front();
    qx.pop();
    int y = qy.front();
    qy.pop();
    vector<pair<int, int> > tmp;
    for (int i = 0; i < 4; i++) {
      int xn = x + dx[i];
      int yn = y + dy[i];
      if (xn >= 0 && xn < n && yn >= 0 && yn < m) {
        if (s[xn][yn] == 'S') continue;
        if (s[xn][yn] == 'T') {
          if (d[xn][yn] > d[x][y] + 1) {
            visited[xn][yn] = 1;
            d[xn][yn] = d[x][y] + 1;
            p[xn][yn][0] = x;
            p[xn][yn][1] = y;
          } else {
            if (d[xn][yn] == d[x][y] + 1) {
              int xx = p[xn][yn][0], yy = p[xn][yn][1];
              if (s[xx][yy] > s[x][y]) {
                p[xn][yn][0] = x;
                p[xn][yn][1] = y;
              }
            }
          }
        } else {
          int c = s[xn][yn] - 'a';
          if ((1 << c) & mask) {
            tmp.push_back(make_pair(xn, yn));
          }
        }
      }
    }
    int sz = tmp.size();
    if (sz) {
      sort(tmp.begin(), tmp.end(), getS);
      for (int i = 0; i < sz; i++) {
        int xn = tmp[i].first;
        int yn = tmp[i].second;
        if (d[xn][yn] > d[x][y] + 1) {
          d[xn][yn] = d[x][y] + 1;
          visited[xn][yn] = true;
          p[xn][yn][0] = x;
          p[xn][yn][1] = y;
          qx.push(xn);
          qy.push(yn);
        } else {
          if (d[xn][yn] == d[x][y] + 1) {
            int xx = p[xn][yn][0], yy = p[xn][yn][1];
            if (s[xx][yy] > s[x][y]) {
              p[xn][yn][0] = x;
              p[xn][yn][1] = y;
            }
          }
        }
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < k; i++) {
    a.push_back(i + 1);
  }
  long long cnt = 0, num = 1, den = 1;
  for (int i = 1; i <= k; i++) {
    num *= 27 - i;
    den *= i;
  }
  cnt = num / den;
  char t;
  for (int i = 0; i < n; i++) {
    scanf("%1c", &t);
    scanf("%s", s[i]);
  }
  int xs = -1, ys = -1, xt = -1, yt = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (s[i][j] == 'S') {
        xs = i;
        ys = j;
      }
      if (s[i][j] == 'T') {
        xt = i;
        yt = j;
      }
    }
  }
  bool was = false;
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < n; j++) {
      for (int h = 0; h < m; h++) {
        visited[j][h] = 0;
        d[j][h] = INF;
        p[j][h][0] = INF;
        p[j][h][1] = INF;
      }
    }
    int mask = 0;
    for (int j = 0; j < k; j++) {
      mask ^= 1 << (a[j] - 1);
    }
    dfs(xs, ys, mask);
    if (visited[xt][yt]) {
      vector<char> res;
      int xc = xt, yc = yt;
      while (!(p[xc][yc][0] == xs && p[xc][yc][1] == ys)) {
        int to1 = p[xc][yc][0];
        int to2 = p[xc][yc][1];
        res.push_back(s[to1][to2]);
        xc = to1;
        yc = to2;
      }
      string cur = "";
      int sz = res.size();
      for (int j = sz - 1; j >= 0; j--) {
        cur += res[j];
      }
      if (!was) {
        ans = cur;
        was = true;
      } else {
        if (cur.length() < ans.length())
          ans = cur;
        else {
          if (cur.length() == ans.length() && cur < ans) ans = cur;
        }
      }
    }
    generate(a, 26);
  }
  if (was) {
    if (ans == "")
      printf("\n");
    else
      cout << ans;
  } else
    printf("-1");
  return 0;
}
