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 static java.util.Arrays.*;
import static java.lang.Math.*;
import static java.math.BigInteger.*;
import java.util.*;
import java.math.*;
import java.io.*;

public class C implements Runnable
{
	String file = "input";
	
	boolean TEST = false;
	
	void solve() throws IOException
	{
		rows = nextInt(); 
		cols = nextInt(); 
		K = nextInt();
		cs = new char[rows][];
		for(int i = 0; i < rows; i++)
		{ 
			cs[i] = next().toCharArray();
			for(int j = 0; j < cols; j++) 
				if(cs[i][j] == 'S')
				{
					sx = i;
					sy = j;
				}
				else if(cs[i][j] == 'T')
				{
					ex = i;
					ey = j;
				}
		}
		
		go(0, 0, 0);
		out.println(res == null ? -1 : res);
	}
	char[][] cs;
	int sx, sy, ex, ey;
	int rows, cols, K;

	void go(int pos, int num, int mask)
	{
		if(num > K) return;
		if(pos == 26)
		{
			get(mask);
			return;
		}
		go(pos + 1, num, mask);
		go(pos + 1, num + 1, mask | (1 << pos));
	}
	int[][] d;
	String res = null;
	void get(int mask)
	{
		BFS(ex, ey, mask);
		if(d[sx][sy] >= INF) return;
		res = min(res, process());
	}
	String min(String a, String b)
	{
		if(a == null) return b;
		return a.compareTo(b) <= 0 ? a : b;
	}
	String process()
	{
		StringBuilder sb = new StringBuilder();
		Queue<Integer> qx = new LinkedList<Integer>();
		Queue<Integer> qy = new LinkedList<Integer>();
		qx.offer(sx);
		qy.offer(sy);
		int dist = d[sx][sy];
		while(dist-- > 1)
		{
			char best = 'z';
			Queue<Integer> qa = new LinkedList<Integer>();
			Queue<Integer> qb = new LinkedList<Integer>();
			boolean[][] vs = new boolean[rows][cols];
			while(!qx.isEmpty())
			{
				int cx = qx.poll(), cy = qy.poll();
				
				for(int k = 0; k < 4; k++)
				{
					int tx = cx + dx[k];
					int ty = cy + dy[k];
					
					if(tx < 0 || tx >= rows || ty < 0 || ty >= cols) continue;
					if(vs[tx][ty]) continue;
					vs[tx][ty] = true;
					if(d[tx][ty] != dist) continue;
					
					qa.offer(tx);
					qb.offer(ty);
					if(cs[tx][ty] < best) best = cs[tx][ty];
				}
			}
			sb.append(best);

			while(!qa.isEmpty())
			{
				int x = qa.poll(), y = qb.poll();
				
				if(cs[x][y] == best)
				{
					qx.offer(x);
					qy.offer(y);
				}
			}
		}
		return sb.toString();
	}
	int[] dx = {-1, 0, 1, 0};
	int[] dy = {0, 1, 0, -1};
	int INF = Integer.MAX_VALUE / 3;
	void BFS(int x, int y, int mask)
	{
		Queue<Integer> qx = new LinkedList<Integer>();
		Queue<Integer> qy = new LinkedList<Integer>();
		qx.offer(x);
		qy.offer(y);
		d = new int[rows][cols];
		for(int i = 0; i < rows; i++) for(int j = 0; j < cols; j++) d[i][j] = INF;
		d[x][y] = 0;
		while(!qx.isEmpty())
		{
			int cx = qx.poll(), cy = qy.poll();
			
			for(int k = 0; k < 4; k++)
			{
				int tx = cx + dx[k];
				int ty = cy + dy[k];
				
				if(tx < 0 || tx >= rows || ty < 0 || ty >= cols) continue;
				if(d[tx][ty] != INF) continue;
				int val = cs[tx][ty] - 'a';
				
				if(cs[tx][ty] == 'S' || (mask >> val & 1) != 0)
				{
					d[tx][ty] = d[cx][cy] + 1;
					qx.offer(tx);
					qy.offer(ty);
				}
			}
		}
	}
	
	String next() throws IOException
	{
		while(st == null || !st.hasMoreTokens()) st = new StringTokenizer(input.readLine());
		return st.nextToken();
	}
	
	int nextInt() throws IOException
	{
		return Integer.parseInt(next());
	}
	
	long nextLong() throws IOException
	{
		return Long.parseLong(next());
	}
	
	double nextDouble() throws IOException
	{
		return Double.parseDouble(next());
	}
	
	void print(Object... o)
	{
		System.out.println(deepToString(o));
	}
	
	void gcj(Object o)
	{
		String s = String.valueOf(o);
		out.println("Case #" + test + ": " + s);
		System.out.println("Case #" + test + ": " + s);
	}
	
	BufferedReader input;
	PrintWriter out;
	StringTokenizer st;
	int test;
	
	void init() throws IOException
	{
		if(TEST) input = new BufferedReader(new FileReader(file + ".in")); 
		else input = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(new BufferedOutputStream(System.out));
	}
	
	public static void main(String[] args) throws IOException
	{
		new Thread(null, new C(), "", 1 << 20).start();
	}
	
	public void run()
	{
		try
		{
			init();
			if(TEST) 
			{
				int runs = nextInt();
				for(int i = 0; i < runs; i++) solve();
			}
			else solve();
			out.close();		
		}
		catch(Exception e)
		{
			e.printStackTrace();
			System.exit(1);
		}
	}
}