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 <iostream>
#include <stdio.h>
#include <memory.h>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>

using namespace std;

#define FOR(i,N) for(int (i)=0;i<N;i++)
#define s(i) scanf("%d",&(i))
#define MP(A,B) pair<int,int>(A,B)

typedef set<char>::iterator  SI ;

set<char> s;
vector<char> v;

char ch[51][51];
bool visited[50][50];
int before[50][50];
int N;
int M;
int K;

pair<int,int> begin;
pair<int,int> end;

string solution;

bool valid[256];

struct Data
{
	int a;
	int b;
	int len;
	char str;
	int back;
	
	Data(int x, int y, int l, int k, char  help)
	{
		a=x;
		b=y;
		str =help;
		back = k;
		len =l;
	}
	
	bool operator < (const Data& d) const
	{
		if(len !=d.len)
		{
			return len>d.len;
		}
		else
		{
			return str>d.str;
		}
	}
};

bool isValid(int a, int b)
{
	return a>=0 && a<N && b>=0 && b<M;
}

string solve()
{
		priority_queue< Data> q;
		q.push( Data(begin.first, begin.second,0,0,' '));

		while(!q.empty())
		{
			Data d = q.top();
			q.pop();
			
			if(!visited[d.a][d.b])
			{
				before[d.a][d.b]=d.back;
				
								
				int x[]={0,-1,1,0};
				int y[]={-1,0,0,1};
				
				
				if(d.a==end.first && d.b== end.second)
				{
					string ret;
					int index = before[d.a][d.b];
					d.a -=x[index];
					d.b -=y[index];
					
					while(!(d.a==begin.first && d.b==begin.second))
					{
						ret+=ch[d.a][d.b];
						index = before[d.a][d.b];
						d.a -=x[index];
						d.b -=y[index];
					}
					
					string res ;
					
					for(int n=ret.length()-1;n>=0;n--)
					{
						res+=ret[n];
					}
					return res;
				}
				
				visited[d.a][d.b]=true;

				for(int n=0;n<4;n++)
				{
					int a=d.a+x[n];
					int b=d.b+y[n];
					if(isValid(a,b) && !visited[a][b])
					{
						q.push(Data(a,b,d.len+1,n,ch[a][b]));
					}
				}
			}
		}
		
		return "-1";
}

void REC(int start, int rest)
{
	if(rest>0)
	{
		for(unsigned int n=start;n<v.size();n++)
		{
			valid[(int)v[n]]=true;
			
			REC(n+1,rest-1);
			
			valid[(int)v[n]]=false;
		}
	}
	else
	{

		memset(visited,false,sizeof(visited));
		
		FOR(n,N)
		{
			FOR(m,M)
			{
				if(!valid[(int)ch[n][m]])
				{
					visited[n][m]=true;
				}
			}
		}
				
				
		string sol = solve();
		
		if(solution == "-1")
		{
			solution =sol;
		}
		else if(sol!="-1")
		{
			if(sol.size()<solution.size() || (sol.size()==solution.size() && sol<solution))
			{
				solution =sol;
			}
		}
				
			
	}
}



int main()
{
	s(N);
	s(M);
	s(K);
	
	FOR(n,N)
	{
		scanf("%s",ch[n]);
	}
	

	
	
	FOR(n,N)
	{
		FOR(m,M)
		{
			if(ch[n][m]=='S')
				begin = MP(n,m);
			else if(ch[n][m]=='T')
				end = MP(n,m);
			else
				s.insert(ch[n][m]);
		}
	}

	
	for(SI it =s.begin();it!=s.end();it++)
	{
		v.push_back(*it);
	}
	
	if((int)v.size()<=K)
	{
		// just go the notmal way
		
		memset(visited,false,sizeof(visited));
		
		string sol = solve();
		
		cout<<sol<<endl;
	}
	else
	{

		solution = "-1";
		memset(valid,false,sizeof(valid));
		
		valid[(int)'S']=true;
		valid[(int)'T']=true;
		
		REC(0,K);
		
		cout<<solution<<endl;
		
		
		
	}
	
	
}