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 java.io.*;

public class Main{

    BufferedReader in;
    StringTokenizer str = null;
    PrintWriter out;

    private String next() throws Exception{
    	while (str == null || !str.hasMoreElements())
    	    str = new StringTokenizer(in.readLine());
    	return str.nextToken();
    }
    
    private int nextInt() throws Exception{
		return Integer.parseInt(next());
    }
    int n,m,k;
    char [][]a;

    String ans = "";

    int []c;

    int sx, sy, tx, ty;

    public void run() throws Exception{
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);
    	
    	n = nextInt();
    	m = nextInt();
    	k = nextInt();
    	
    	a = new char[n][m];
    	for(int i = 0; i < n; ++i) a[i] = next().toCharArray();

    	sx = -1;
    	sy = -1;
    	tx = -1;
    	ty = -1;

    	for(int i = 0; i < n; ++i) {
    		for(int j = 0; j < m; ++j) {
    			if (a[i][j] == 'S') {
    				sx = i;
    				sy = j;
    			}

    			if (a[i][j] == 'T') {
    				tx = i;
    				ty = j;
    			}
    		}
    	}

    	c = new int[k];
    	C(-1, 0);
    	if (ans.length() > 0) {
    		out.println(ans.substring(1, ans.length()-1));
    	}else {
    		out.println(-1);
    	}
        out.close();
    }

    private void C(int start, int at) {
    	if (at == c.length) {
    		String path = "";
    		for(int i = 0; i < at; ++i) path+=(char)(c[i] + 'a');
    		// System.out.println(path);
    		String tans = bfs(path, tx, ty, sx, sy);

    		if (!tans.equals("")) {
    			// System.out.println(tans + " " + path);
    			if (ans.equals("") || ans.length() > tans.length() || (ans.length() == tans.length() && ans.compareTo(tans) > 0)) {
    				ans = tans;
    			}
    		}
    		return;
    	}
    	for(int i = start + 1; i < 26; ++i) {
    		c[at] = i;
    		C(i, at + 1);
    	}
    }

    final static int oo = Integer.MAX_VALUE/2;
    final static int BASE = 50;

    int di[] = new int[]{-1, 1, 0, 0};
    int dj[] = new int[]{0, 0, -1, 1};

    Queue<Integer> q;
    int [][]d;

    StringBuilder sb;

    private String bfs(String path, int sx, int sy, int tx, int ty) {
    	d = new int[n][m];
    	for(int i = 0; i < n; ++i) Arrays.fill(d[i], oo);
    	d[sx][sy] = 0;

    	q = new ArrayDeque<Integer>();
    	q.add(sx * BASE + sy);

    	while(q.size() > 0) {
    		int cur = q.poll();
    		int x = cur / BASE, y = cur % BASE;
    		for(int i = 0; i < di.length; ++i) {
    			int nx = x + di[i], ny = y + dj[i];
    			add(nx, ny, x, y, path);
    		}
    	}

    	if (d[tx][ty] == oo) return "";
    	sb = new StringBuilder();
    	build(tx, ty, d[tx][ty]);
    	// return sb.reverse().toString();
    	return sb.toString();
    }

    private void build(int x, int y, int dis) {
    	sb.append(a[x][y]);
    	if (dis == 0) return;
    	char move = ' ';
    	int idx = -1;
    	for(int i = 0; i < di.length; ++i) {
    		int nx = x + di[i], ny = y + dj[i];
    		if (validMove(nx, ny, x, y)) {
    			if (move == ' ' || move > a[nx][ny]) {
    				move = a[nx][ny];
    				idx = i;
    			}
    		}
    	}

    	build(x + di[idx], y + dj[idx], dis - 1);
    }

    private boolean validMove(int x, int y, int fx, int fy) {
    	if (x < 0 || y < 0) return false;
    	if (x >= n || y >= m) return false;
    	return d[x][y] + 1 == d[fx][fy];
    }

    private void add(int x, int y, int fx, int fy, String path) {
    	if (x < 0 || y < 0) return;
    	if (x >= n || y >= m) return;
    	if (a[x][y] != 'S' && path.indexOf(a[x][y]) == -1) return;
    	if (d[x][y] > d[fx][fy] + 1) {
    		d[x][y] = d[fx][fy] + 1;
    		q.add(x * BASE + y);
    	}
    }

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