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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Set;
import java.util.StringTokenizer;

public class E {

	static class State implements Comparable<State> {
		int x, y, used, rem, dist;
		State prev;
		char type;

		State(State prev, int x, int y, int used, int rem, char type, int dist) {
			this.prev = prev;
			this.x = x;
			this.y = y;
			this.used = used;
			this.rem = rem;
			this.type = type;
			this.dist = dist;
		}

		@Override
		public int compareTo(State o) {
			return dist == o.dist ? o.rem - rem : dist - o.dist;
		}
	}

	public static void main(String[] args) throws IOException {
		String letters = "";
		for (char c = 'a'; c <= 'z'; c++) {
			letters += c;
		}
		letters += 'S';

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int m = Integer.parseInt(st.nextToken());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		int sx = -1;
		int sy = -1;
		int ex = -1;
		int ey = -1;
		Map<Integer, Set<Integer>> v = new HashMap<Integer, Set<Integer>>();
		int[][] g = new int[m][n];
		for (int i = 0; i < m; i++) {
			String line = br.readLine();
			for (int j = 0; j < n; j++) {
				char c = line.charAt(j);
				if (c == 'S') {
					sx = i;
					sy = j;
				} else if (c == 'T') {
					ex = i;
					ey = j;
				}
				g[i][j] = letters.indexOf(c);
			}
		}

		int[] dx = { -1, 0, 1, 0 };
		int[] dy = { 0, 1, 0, -1 };

		int neiCount;
		int[] nei = new int[4];
		int[] dir = new int[4];

		PriorityQueue<State> q = new PriorityQueue<State>();
		q.add(new State(null, sx, sy, 0, k, '\0', Math.abs(ex - sx)
				+ Math.abs(ey - sy)));
		State res = null;
		while (!q.isEmpty()) {
			State cur = q.poll();
			if (cur.x == ex && cur.y == ey) {
				res = cur;
				break;
			}
			neiCount = 0;
			for (int i = 0; i < 4; i++) {
				int tx = cur.x + dx[i];
				if (tx < 0 || tx == m)
					continue;
				int ty = cur.y + dy[i];
				if (ty < 0 || ty == n)
					continue;
				if (tx == sx && ty == sy)
					continue;
				nei[neiCount] = g[tx][ty];
				dir[neiCount++] = i;
			}
			for (int i = 0; i < neiCount; i++) {
				for (int j = neiCount - 1; j > i; j--) {
					if (nei[j] < nei[j - 1]) {
						int t = nei[j];
						nei[j] = nei[j - 1];
						nei[j - 1] = t;
						t = dir[j];
						dir[j] = dir[j - 1];
						dir[j - 1] = t;
					}
				}
			}
			if (neiCount > 0 && nei[0] < 0) {
				res = new State(cur, cur.x + dx[dir[0]], cur.y + dy[dir[0]], 0,
						0, 'T', 0);
				break;
			}
			for (int i = 0; i < neiCount; i++) {
				boolean alreadyUsed = ((1 << nei[i]) & cur.used) != 0;
				if (!alreadyUsed && cur.rem == 0)
					continue;
				int newUsed = (1 << nei[i]) | cur.used;
				int tx = cur.x + dx[dir[i]];
				int ty = cur.y + dy[dir[i]];
				int blah = (tx << 6) | ty;
				if (!v.containsKey(newUsed))
					v.put(newUsed, new HashSet<Integer>());
				if (!v.get(newUsed).contains(blah)) {
					v.get(newUsed).add(blah);
					q.add(new State(cur, tx, ty, newUsed, alreadyUsed ? cur.rem
							: cur.rem - 1, (char) ('a' + nei[i]), Math.abs(tx
							- ex)
							+ Math.abs(ty - ey)));
				}
			}
		}
		if (res == null) {
			System.out.println(-1);
		} else {
			StringBuilder sb = new StringBuilder();
			print(res.prev, sb);
			System.out.println(sb);
		}
	}

	private static void print(State s, StringBuilder sb) {
		if (s.prev == null)
			return;
		sb.insert(0, s.type);
		print(s.prev, sb);
	}

}
