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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <ctime>
#define MAXN 55
#define ANY '*'
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
typedef long long LL;
typedef pair<int, int> pii;

char g[MAXN][MAXN];
int n, m, k;
pii st, end;
bool flag[128], yes[128];
string final;
int dis[MAXN][MAXN];
const int dx[] = { 0, 0, 1, -1 };
const int dy[] = { 1, -1, 0, 0 };
bool got;

bool inq[MAXN][MAXN];
pii myQ[2][100000];

void chk() {
	queue<pii> q;
	memset(dis, 0x3f, sizeof(dis));
	q.push(end);
	dis[end.first][end.second] = 0;
	while (!q.empty()) {
		pii cur = q.front();
		int next = dis[cur.first][cur.second] + 1;
		q.pop();
		for (int d = 0; d < 4; ++d) {
			int tx = cur.first + dx[d];
			int ty = cur.second + dy[d];
			if (flag[(int) g[tx][ty]] && dis[tx][ty] == INF) {
				dis[tx][ty] = next;
				q.push(make_pair(tx, ty));
			}
		}
	}

	if (dis[st.first][st.second] == INF)
		return;

	int sz[2]={ }, curQ = 0, preQ = 1, OOXX = dis[st.first][st.second];
	string t = "";
	memset(inq, 0, sizeof(inq));
	myQ[curQ][sz[curQ]++] = st;
	inq[st.first][st.second] = true;
	for (int ooxx = 1; ooxx < OOXX; ++ooxx) {
		swap(curQ, preQ);
		sz[curQ] = 0;

		char c = 'z';
		for (int i = 0; i < sz[preQ]; ++i) {
			pii cur = myQ[preQ][i];
			inq[cur.first][cur.second] = false;
			for (int d = 0; d < 4; ++d) {
				int tx = cur.first + dx[d];
				int ty = cur.second + dy[d];
				if (dis[tx][ty] == dis[cur.first][cur.second] -1
						&& inq[tx][ty] == false) {
					inq[tx][ty] = true;
					myQ[curQ][sz[curQ]++] = make_pair(tx, ty);
					c = min(c, g[tx][ty]);
				}
			}
		}

		for (int i = 0; i < sz[curQ]; ++i) {
			pii cur = myQ[curQ][i];
			if (g[cur.first][cur.second] != c) {
				myQ[curQ][i] = myQ[curQ][sz[curQ] - 1];
				--sz[curQ];
				--i;
			}
		}

		t += c;
	}

	if(!got) {
		final=t;
		got=true;
	} else
		final = min(final, t);
}

void dfs(int p, int dep) {
	chk();
	if (dep >= k)
		return ;
	for (int i = p; i <= 'z'; ++i) {
		if (!yes[i])
			continue;
		flag[i] = true;
		dfs(i + 1, dep + 1);
		flag[i] = false;
	}
}

int main() {
#ifndef ONLINE_JUDGE
//	freopen("in", "r", stdin);
#endif

	while (scanf("%d %d %d", &n, &m, &k) != EOF) {
		memset(g, 0, sizeof(g));
		memset(yes, 0, sizeof(yes));
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= m; ++j) {
				scanf(" %c", &g[i][j]);
				if (g[i][j] == 'S') {
					g[i][j] = ANY;
					st = make_pair(i, j);
				} else if (g[i][j] == 'T') {
					g[i][j] = ANY;
					end = make_pair(i, j);
				}
				yes[(int) g[i][j]] = true;
			}
		}

		final = "";
		got=false;
		memset(flag, 0, sizeof(flag));
		flag[ANY] = true;
		dfs('a', 0);
		if (!got)
			puts("-1");
		else
			cout << final << endl;
	}

	return 0;
}
