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

import java.util.*;

import static java.lang.Math.*;

public class C {
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		N = in.nextInt();
		M = in.nextInt();
		K = in.nextInt();
		B = new char[N][];
		for(int i = 0; i < N;i++)
			B[i] = in.next().toCharArray();
		best = null;
		for(int r = 0; r < N;r++){
			for(int c = 0; c < M;c++){
				if(B[r][c] == 'S'){
					sr = r;
					sc = c;
				}
				if(B[r][c] == 'T'){
					er = r;
					ec = c;
				}
			}
		}
		dist = new int[N][M];
		done = new boolean[N][M];
		recur(-1, 0, new boolean[26]);
		if(best == null)
			System.out.println("-1");
		else
			System.out.println(best);
	}
	private static void recur(int prev, int total, boolean[] used) {
		if(total == K){
			solve(used);
			return;
		}
		for(int i = prev+1; i < 26;i++){
			used[i] = true;
			recur(i, total+1, used);
			used[i] = false;
		}
	}
	private static void solve(boolean[] U) {
		for(int[] d:dist)
			Arrays.fill(d, Integer.MAX_VALUE);
		dist[er][ec] = 0;
		LinkedList<Integer> Q = new LinkedList<Integer>();
		Q.add(er*M+ec);
		while(Q.size() > 0){
			int a = Q.poll();
			int r = a/M;
			int c = a%M;
			for(int d = 0; d < 4;d++){
				int nr = r + dr[d];
				int nc = c + dc[d];
				if(nr < 0 || nr >= N || nc < 0 || nc >= M)
					continue;
				if(B[nr][nc] != 'S' && B[nr][nc] != 'T' && !U[B[nr][nc]-'a'])
					continue;
				if(dist[nr][nc] == Integer.MAX_VALUE){
					dist[nr][nc] = dist[r][c] + 1;
					Q.add(nr*M+nc);
				}
			}
		}
		if(dist[sr][sc] == Integer.MAX_VALUE || (best != null && dist[sr][sc]-1 > best.length()))
			return;
		String ans = "";
		Q = new LinkedList<Integer>();
		
		Q.add(sr*M+sc);
		for(boolean[] b:done)
			Arrays.fill(b, false);
		while(true){
			char bestch = 'z';
			LinkedList<Integer> NQ = new LinkedList<Integer>();
			while(Q.size() > 0){
				int a = Q.poll();
				int r = a/M;
				int c = a%M;
				for(int d = 0; d < 4;d++){
					int nr = r + dr[d];
					int nc = c + dc[d];
					if(nr < 0 || nr >= N || nc < 0 || nc >= M)
						continue;
					if(B[nr][nc] != 'S' && B[nr][nc] != 'T' && !U[B[nr][nc]-'a'])
						continue;
					if(dist[nr][nc] != dist[r][c]-1)
						continue;
					if(done[nr][nc])
						continue;
					done[nr][nc] = true;
					if(bestch == B[nr][nc]){
						NQ.add(nr*M+nc);
					}else if(B[nr][nc] == 'T' || (bestch > B[nr][nc] && bestch != 'T')){
						NQ.clear();
						NQ.add(nr*M+nc);
						bestch = B[nr][nc];
					}
				}
			}
			if(bestch == 'T')
				break;
			ans += bestch;
			Q = NQ;
		}
		if(best == null || ans.length() < best.length() || ans.compareTo(best) < 0){
			best = ans;
		}
	}
	static boolean[][] done;
	static int[] dr = {-1, 1, 0, 0};
	static int[] dc = {0, 0, -1, 1};
	static int[][] dist;
	static int sr, sc, er, ec;
	static String best;
	static int N, M, K;
	static char[][] B;
}
