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

//Abra at 13.05.2011 19:29:36
import java.io.*;
import java.math.*;
import java.util.*;

public class CF72_C {
	int[][] map, pa;
	int n, m, k;
	TreeSet<Integer> al;
	int xs = -1, ys = -1, xt = -1, yt = -1;
	TreeSet<Integer> av;
	
	void solve() throws IOException {
		n = nextInt();
		m = nextInt();
		k = nextInt();
		if (n * m == 2) {
			return;
		}
		map = new int[n][m];
		br.readLine();
		al = new TreeSet<Integer>();
		for (int i = 0; i < n; i++) {
			String s = br.readLine();
			for (int j = 0; j < m; j++)
				if (s.charAt(j) == 'S') {
					map[i][j] = -2;
					xs = i;
					ys = j;
				} else
					if (s.charAt(j) == 'T') {
						map[i][j] = -1;
						xt = i;
						yt = j;
					} else {
						map[i][j] = s.charAt(j);
						al.add(map[i][j]);
					}
		}
		Vector<Integer> v = new Vector<Integer>();
		for (Integer t : al)
			v.add(t);
		HashSet<TreeSet<Integer>> per = new HashSet<TreeSet<Integer>>();
		Random R = new Random();
		for (int i = 0; i < 10000; i++) {
			TreeSet<Integer> ad = new TreeSet<Integer>();
			for (int j = 0; j < k; j++) {
				ad.add(v.get(R.nextInt(v.size())));
			}
			if (ad.size() != Math.min(al.size(), k)) continue;
			per.add(ad);
		}		
		String p = "";
		for (TreeSet<Integer> ts : per) {
			av = ts;
			pa = new int[n][m];
			av.add(-2);
			wave(xt, yt, 1);
			if (pa[xs][ys] == 0) continue;
			TreeSet<Integer> q = new TreeSet<Integer>();
			q.add(xs * 50 + ys);
			cm = "";
			while (true) {
				TreeSet<Integer> qq = new TreeSet<Integer>();
				int mm = 10000;
				for (Integer t : q) {
					int x = t / 50;
					int y = t % 50;
					if (x > 0 && pa[x - 1][y] == pa[x][y] - 1 && pa[x - 1][y] != 0) {
						if (map[x - 1][y] < mm) {
							mm = map[x - 1][y];
							qq.clear();
						}
						if (map[x - 1][y] == mm) {
							qq.add((x - 1) * 50 + (y));
						}
					}
					if (x < n - 1 && pa[x + 1][y] == pa[x][y] - 1 && pa[x + 1][y] != 0) {
						if (map[x + 1][y] < mm) {
							mm = map[x + 1][y];
							qq.clear();
						}
						if (map[x + 1][y] == mm) {
							qq.add((x + 1) * 50 + (y));
						}
					}
					if (y > 0 && pa[x][y - 1] == pa[x][y] - 1 && pa[x][y - 1] != 0) {
						if (map[x][y - 1] < mm) {
							mm = map[x][y - 1];
							qq.clear();
						}
						if (map[x][y - 1] == mm) {
							qq.add((x) * 50 + (y - 1));
						}
					}
					if (y < m - 1 && pa[x][y + 1] == pa[x][y] - 1 && pa[x][y + 1] != 0) {
						if (map[x][y + 1] < mm) {
							mm = map[x][y + 1];
							qq.clear();
						}
						if (map[x][y + 1] == mm) {
							qq.add((x) * 50 + (y + 1));
						}
					}
				}
				if (qq.size() == 0) break;
				if (mm != -1) cm += (char) mm;
				q = qq;
			}
			if ((cm.compareTo(p) < 0 && p.length() >= cm.length() && pa[xs][ys] == cm.length() - 1) || p.length() == 0) p = cm;
		}
		if (p.length() == 0) out.print(-1); else
		out.println(p);
	}
	
	String cm = "";
	//TreeMap<>
	
	void wave(int x, int y, int z) {
		pa[x][y] = z;
		if (x > 0 && av.contains(map[x - 1][y]) && (pa[x - 1][y] == 0 || map[x - 1][y] == -2 || pa[x - 1][y] > z + 1)) {
			wave(x - 1, y, z + 1);
		}
		if (y > 0 && av.contains(map[x][y - 1]) && (pa[x][y - 1] == 0 || map[x][y - 1] == -2 || pa[x][y - 1] > z + 1)) {
			wave(x, y - 1, z + 1);
		}
		if (x < n - 1 && av.contains(map[x + 1][y]) && (pa[x + 1][y] == 0 || map[x + 1][y] == -2 || pa[x + 1][y] > z + 1)) {
			wave(x + 1, y, z + 1);
		}
		if (y < m - 1 && av.contains(map[x][y + 1]) && (pa[x][y + 1] == 0 || map[x][y + 1] == -2 || pa[x][y + 1] > z + 1)) {
			wave(x, y + 1, z + 1);
		}
	}

	public static void main(String[] args) throws IOException {
		new CF72_C().run();
	}

	StreamTokenizer in;
	PrintWriter out;
	boolean oj;
	BufferedReader br;

	void init() throws IOException {
		oj = System.getProperty("ONLINE_JUDGE") != null;
		Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("input.txt");
		Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("output.txt");
		br = new BufferedReader(reader);
		in = new StreamTokenizer(br);
		out = new PrintWriter(writer);
	}

	long beginTime;

	void run() throws IOException {
		beginTime = System.currentTimeMillis();
		long beginMem = Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory();
		init();
		solve();
		long endMem = Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory();
		long endTime = System.currentTimeMillis();
		if (!oj) {
			System.out.println("Memory used = " + (endMem - beginMem));
			System.out.println("Total memory = " + (Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()));
			System.out.println("Running time = " + (endTime - beginTime));
		}
		out.flush();
	}

	int nextInt() throws IOException {
		in.nextToken();
		return (int) in.nval;
	}

	long nextLong() throws IOException {
		in.nextToken();
		return (long) in.nval;
	}

	String nextString() throws IOException {
		in.nextToken();
		return in.sval;
	}

	double nextDouble() throws IOException {
		in.nextToken();
		return in.nval;
	}
}
