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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <cmath>
#include <cstring>
#include <string>
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef vector<int> VI;
typedef pair<int,int> PII;
 
#define REP(i,n) for(int i=0;i<(n);++i)
#define SIZE(c) ((int)((c).size()))
#define FOR(i,a,b) for (int i=(a); i<(b); ++i)
#define FOREACH(i,x) for (__typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define FORD(i,a,b) for (int i=(a)-1; i>=(b); --i)
#define ALL(v) (v).begin(), (v).end()
 
#define pb push_back
#define mp make_pair
#define st first
#define nd second

char level[100][100];

struct state{
  PII pos;
  int mask;
  int d;
  state(PII p, int m, int di = 0):pos(p),mask(m),d(di){}
  state(){}
};

struct Cmp{
bool operator()(const state& s1, const state& s2){
    if(s1.mask != s2.mask)return s1.mask<s2.mask;
    return s1.pos < s2.pos;
}
};

int bitc[1<<13];
int bitcount(int i){
  return bitc[i>>13] + bitc[i%(1<<13)];
};

int main(){
  REP(i,1<<13){
    int j = i;
    while(j){
      bitc[i] += j&1;
      j>>=1;
    }
  }

  int N,M,K;
  scanf("%d%d%d",&N,&M,&K);
  REP(i,N)
    scanf("%s",level[i]);
  PII S, T;
  REP(i,N)REP(j,M){
    if(level[i][j] == 'S')S=PII(i,j);
    else if(level[i][j] == 'T')T=PII(i,j);
    else level[i][j] -= 'a';
  }
  map<state, state, Cmp> visited;
  queue<state> Q;
  Q.push(state(S,0,0));
  visited[state(S,0,0)] = state();
  while(!Q.empty()){
    PII pos = Q.front().pos;
    int d = Q.front().d;
    int mask = Q.front().mask;
    state s(pos,mask,d);
    Q.pop();        
    vector<pair<char,PII> > poses;
    if(pos.st>0)poses.pb(mp(level[pos.st-1][pos.nd], PII(pos.st-1,pos.nd)));
    if(pos.st<N-1)poses.pb(mp(level[pos.st+1][pos.nd], PII(pos.st+1,pos.nd)));
    if(pos.nd>0)poses.pb(mp(level[pos.st][pos.nd-1], PII(pos.st,pos.nd-1)));    
    if(pos.nd<M-1)poses.pb(mp(level[pos.st][pos.nd+1], PII(pos.st,pos.nd+1)));
    sort(poses.begin(), poses.end());
    
    REP(i,poses.size()){
      PII& pos2 = poses[i].nd;
      if(pos2 == T){
        stack<char> stos;
        while(s.pos != S){
          stos.push(level[s.pos.st][s.pos.nd]+'a');
          s = visited[s];
        }
        while(!stos.empty()){printf("%c",stos.top());stos.pop();}
        printf("\n");
        return 0;
      }
      int mask2 = mask | (1<<(level[pos2.st][pos2.nd])); 
      state s2(pos2,mask2,d+1);
      if(bitcount(mask2) <= K && visited.find(s2) == visited.end()){
        visited[s2] = s;
        Q.push(s2);
      }    
    }
  }
  printf("-1\n");
}