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());
    }
    
    char [][]a;
    int n, m, k;
    int [][][]d;
    
    final static int oo = Integer.MAX_VALUE/2;
    Queue<S> q;
    
    int di[] = new int[]{-1, 1, 0, 0};
    int dj[] = new int[]{0, 0, -1, 1};
    
    StringBuilder path;
    
    public void run() throws Exception{
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);
	n = nextInt();
	m = nextInt();
	k = nextInt() + 2;
	a = new char[n][m];
	for(int i = 0; i < n; ++i) a[i] = next().toCharArray();
	
	int 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;
		}
	    }
	}
	
	d = new int[n][m][k];
	for(int i = 0; i < n; ++i){
	    for(int j = 0; j < m; ++j){
		Arrays.fill(d[i][j], oo);
	    }
	}
	
	d[sx][sy][0] = 0;
	q = new PriorityQueue<S>();
	q.add(new S(sx, sy, 0, 0));
	
	while(q.size() > 0){
	    S cur = q.poll();
	    //System.out.println(cur);
	    for(int i = 0; i < di.length; ++i){
		int nx = cur.x + di[i], ny = cur.y + dj[i];
		add(nx, ny, cur.x, cur.y, cur.k);
	    }
	}
	
	int nk = -1;
	String ans = "";
	for(int i = 0; i < k; ++i){
	    if (d[tx][ty][i] != oo){
		/*
		if (nk == -1 || d[tx][ty][nk] > d[tx][ty][i]){
		    nk = i;
		}
		*/
		path = new StringBuilder();
		print(tx, ty, i, d[tx][ty][i]);
		String tpath = path.reverse().toString();
		if (ans.equals("") || ans.compareTo(tpath) > 0){
		    ans = tpath;
		}
		//out.println(path);
	    }
	}
	
	//out.println(d[0][1][1]);
	
	if (ans.equals("")){
	    out.println(-1);
	    out.close();
	    return;
	}
	//out.println(d[tx][ty][nk]);
	//out.println("k=" + nk);
	//print(tx, ty, nk, d[tx][ty][nk]);
	out.println(ans.substring(1, ans.length()-1));
        out.close();
    }
    
    private void print(int tx, int ty, int tk, int dis){
	/*if (a[tx][ty] != 'T' && a[tx][ty] != 'S') out.print(a[tx][ty]);*/
	//out.print(a[tx][ty]);
	path.append(a[tx][ty]);
	if (dis == 0) return;
	int cur = d[tx][ty][tk];
	int idx = -1;
	char move = ' ';
	for(int i = 0; i < 4; ++i){
	    int nx = tx + di[i], ny = ty + dj[i];
	    if (ok(nx, ny, tx, ty, tk)){
		if (move == ' ' || move > a[nx][ny]){
		    move = a[nx][ny];
		    idx = i;
		}
	    }
	}
	
	int nk = tk;
	int nx = tx + di[idx], ny = ty + dj[idx];
	if (a[nx][ny] != a[tx][ty]) nk--;
	
	print(nx, ny, nk, dis - 1);
    }
    
    private boolean ok(int x, int y, int fx, int fy, int fk){
	if (x < 0 || y < 0) return false;
	if (x >= n || y >= m) return false;
	int nk = fk;
	if (a[x][y] != a[fx][fy]) nk--;
	if (nk < 0) return false;
	return d[x][y][nk] + 1 == d[fx][fy][fk];
    }
    
    private void add(int x, int y, int fx, int fy, int fk){
	if (x < 0 || y < 0) return;
	if (x >= n || y >= m) return;
	if (a[x][y] == 'S') return;
	int nk = fk;
	if (a[x][y] != a[fx][fy]) nk++;
	if (nk >= k) return;
	if (d[x][y][nk] > d[fx][fy][fk] + 1){
	    d[x][y][nk] = d[fx][fy][fk] + 1;
	    q.add(new S(x, y, nk, d[x][y][nk]));
	}
    }
    
    class S implements Comparable<S>{
	int x, y, k, d;
	
	public S(int x, int y, int k, int d){
	    this.x = x;
	    this.y = y;
	    this.k = k;
	    this.d = d;
	}
	
	public int compareTo(S other){
	    if (this.d < other.d) return -1;
	    if (this.d > other.d) return 1;
	    //if (this.k < other.k) return -1;
	    //if (this.k > other.k) return 1;
	    return 0;
	}
	
	public String toString(){
	    return this.x + " " + this.y + " " + this.k + " " + this.d;
	}
    }

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