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 <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cfloat>


 
using namespace std;
 
const int oo = 0x3f3f3f3f;
const double eps = 1e-9;

typedef long long ll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
 
#define sz(c) int((c).size())
#define all(c) (c).begin(), (c).end()
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define FORD(i,a,b) for (int i = int(b)-1; i >= (a); i--)
#define FORIT(i,c) for (__typeof__((c).begin()) i = (c).begin(); i != (c).end(); i++)

string feld[60];
vector<pair<char,int> > adj[3000];
int dr[4]={-1,1,0,0};
int dc[4]={0,0,-1,1};
int best;
char way[3000];
int cbest;
char cway[3000];
bool ok[300];
char lastC[3000];
int prev[3000];
int s,e,R,C;
void search(int let, int lid){
	if(let){
		while(lid<='z'){
			ok[lid]=true;
			search(let-1,lid+1);
			ok[lid++]=false;
		}
	} else {
		//FOR(i,0,R*C)cout << i <<  " "<<prev[i]<<endl;
		queue<int> pre = queue<int>();
		queue<int> q = queue<int>();
		q.push(s);
		while(!q.empty()){
			int n = q.front();
			q.pop();
			FOR(i,0,sz(adj[n])){
				if(adj[n][i].first=='T'){
					cbest = 0;
					while(prev[n]!=-1){
						cway[cbest++]=lastC[n];
						n = prev[n];
					}
					if(cbest<best){
						FOR(i,0,cbest)way[i]=cway[cbest-1-i];
						best = cbest;
					} else if(cbest==best){
						bool sm = true;
						FOR(i,0,cbest){
							if(way[i]<cway[cbest-1-i]){
								sm = false;
								break;
							} else if(way[i]>cway[cbest-1-i]){
								break;
							}
						}
						if(sm)FOR(i,0,cbest)way[i]=cway[cbest-1-i];
					}
					while(!pre.empty()){
						prev[pre.front()]=-1;
						pre.pop();
					}
					return;
				} else if(ok[adj[n][i].first]){
					int to = adj[n][i].second;
					if(prev[to]==-1){
						prev[to]=n;
						lastC[to]=adj[n][i].first;
						pre.push(to);
						q.push(to);
					}
				}
			}
		}
					while(!pre.empty()){
						prev[pre.front()]=-1;
						pre.pop();
					}
	}
}
int main(){
	int K;
	cin >> R >> C >> K;
	FOR(i,0,R)cin >> feld[i];
	FOR(i,0,R*C)adj[i].clear();
	FOR(r,0,R)FOR(c,0,C){
		FOR(d,0,4){
			int rr = r+dr[d];
			int cc = c+dc[d];
			if(rr<0||rr==R||cc<0||cc==C)continue;
			adj[r*C+c].push_back(make_pair(feld[rr][cc],rr*C+cc));
		}
		sort(all(adj[r*C+c]));
	}
	s =-1;e = -1;
	FOR(r,0,R)FOR(c,0,C){
		if(feld[r][c]=='S')s = r*C+c;
		else if(feld[r][c]=='T') e = r*C+c;
	}
	best = oo;
	memset(ok,false,sizeof(ok));
	memset(prev,-1,sizeof(prev));
	search(K,'a');
	if(best==oo){
		printf("-1\n");
	} else {
		FOR(i,0,best)printf("%c",way[i]);
		printf("\n");
	}
	return 0;
}
