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 <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const int Inf = 0x3f3f3f3f;

int N, M, K;
char G[110][110];
char dic[110 * 110];
bool use[150];
int dis[110][110];
int ans, sx, sy, ex, ey, Cnt;
string path;
int q[110 * 110];
int move[4][2] = {0, 1, 0, -1, -1, 0, 1, 0};

bool in(int x, int y)
{
	return x >= 0 && x < N && y >= 0 && y < M;	
}

void getPath(int x, int y, string s, int L, bool flag)
{
//	cout << s << endl;
	if(L == ans-1)
	{
		path = s;
		return;	
	}
	
	for(int k=0; k<4; k++)
	{
		int a = x + move[k][0];
		int b = y + move[k][1];
		if(in(a, b) && dis[x][y] == dis[a][b]+1 && (G[a][b] <= path[L] || flag))
		{
			getPath(a, b, s+G[a][b], L+1, flag | G[a][b] < path[L]);	
		}	
	}
}

void bfs()
{
//	puts("output----");
//		for(char i=0; i<='z'; i++)
//			if(use[i])
//				putchar(i);
//		cout << endl;

		
	int l = 0, r = 0;
	q[r++] = ex*M + ey;
	dis[ex][ey] = 0;
	dis[sx][sy] = Inf;
	
	while(l < r)
	{
		int x = q[l] / M;
		int y = q[l++] % M;
		if(dis[x][y] >= ans)	break;
		for(int k=0; k<4; k++)
		{
			int a = x + move[k][0];
			int b = y + move[k][1];
			if(!in(a, b) || dis[a][b] < dis[x][y] + 1)	continue;
			if(G[a][b] == 'S' || use[G[a][b]])	
			{
				dis[a][b] = dis[x][y] + 1;
				if(G[a][b] == 'S')	break;
				q[r ++] = a*M + b;	
			}
		}	
	}
	
	if(dis[sx][sy] <= ans)
	{
//		puts("output");
//		for(int i=0; i<='z'; i++)
//			if(use[i])
//				putchar(i);
//		cout << endl;

		if(dis[sx][sy] < ans)
		{
			ans = dis[sx][sy];
			path.clear();
			for(int i=0; i<ans-1; i++)
			{
				path.push_back('z');
			}
		}
		getPath(sx, sy, "", 0, 0);	
	}
	
	for(int i=0; i<r; i++)
	{
		dis[q[i] / M][q[i] % M] = Inf;	
	}
}

void dfs(int k, int st)
{
	if(k == Cnt || k == K)
	{
		bfs();	
		return;
	}
	
	while(++st < Cnt)
	{
		use[dic[st]] = 1;
		dfs(k+1, st);
		use[dic[st]] = 0;	
	}
}


int main()
{
	while(cin >> N >> M >> K)
	{
		Cnt = 0;
		for(int i=0; i<N; i++)
		{
//			scanf("%s", G[i]);
			for(int j=0; j<M; j++)
			{
				G[i][j] = 'a';
				if(G[i][j] == 'S')	sx = i, sy = j;
				else if(G[i][j] == 'T')	ex = i, ey = j;
				else dic[Cnt++] = G[i][j];	
			}	
		}
		sx = sy = 0;
		ex = N-1, ey = M-1;
		
		sort(dic, dic + Cnt);
		Cnt = unique(dic, dic + Cnt) - dic;
		
//		for(int i=0; i<Cnt; i++)	putchar(dic[i]);cout << endl;
		ans = N * M;
		memset(use, 0, sizeof(use));
		memset(dis, 0x3f, sizeof(dis));
		dfs(0, -1);
		
		if(ans == N * M)	puts("-1");
		else	cout << path << endl;
	}
	
	
	return 0;	
}
