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

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector") 
#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
 
#include <bits/stdc++.h>
 
#define int long long
#define ll long long
#define ull unsigned long long
#define ld long double
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define repb(i, r, l) for (int i = r; i >= l; i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define ret(a) {cout << a; exit(0);}
 
using namespace std;
 
mt19937 mrand(random_device{}()); 
const ll mod = 1e9 + 7;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
 
typedef pair < int , int > pii ;                                            
typedef pair < long long  , long long > pll ;
const int N = 2e5 + 5, K = 2e9 + 1;
 
int n, m , k;

string s = "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz";
char ch[55][55];
int a[55][55];
int ans = 1e9;
int d[55][55];
string S[55][55];
int xe,ye,xs,ys;
string dfs(int x, int y){
	if(ch[x][y] == 'T')
		return S[x][y];
	if(sz(S[x][y]) != 0)
		return S[x][y];
	S[x][y] = 'z';
	if(x+1<= n and d[x][y] == d[x+1][y]+1)
		S[x][y] = min(S[x][y] , dfs(x+1,y));
	if(x-1> 0 and d[x][y] == d[x-1][y]+1)
		S[x][y] = min(S[x][y], dfs(x-1,y));
	if(y+1 <= m and d[x][y] == d[x][y+1]+1)
		S[x][y] = min(S[x][y],dfs(x,y+1));
	if(y-1> 0 and d[x][y] == d[x][y-1]+1)
		S[x][y] = min(S[x][y], dfs(x,y-1));
	if(ch[x][y] != 'S')
		S[x][y] = ch[x][y] + S[x][y];
	return S[x][y];
}
void check(int mask){
	for(int i = 1; i <= n;i++)
		for(int j = 1 ; j <= m; j++)
			d[i][j] = -1,S[i][j].clear();
	deque < pii > dq;
	d[xe][ye] = 0;
	dq.pb({xe,ye});
	while(sz(dq) != 0){
		pii tmp = dq.front();dq.pop_front();
		int x = tmp.fi , y = tmp.se;
		if(x + 1 <= n and d[x+1][y] == -1 and (a[x+1][y]&mask)==a[x+1][y])
			d[x+1][y] = d[x][y] + 1,dq.pb({x+1,y});
		if(x-1>0 and d[x-1][y] == -1 and (a[x-1][y]&mask)==a[x-1][y])
			d[x-1][y] = d[x][y] + 1,dq.pb({x-1,y});			
		if(y+1 <= m and d[x][y+1] == -1 and (a[x][y+1]&mask)==a[x][y+1])
			d[x][y+1] = d[x][y] + 1, dq.pb({x,y+1});
		if(y-1 > 0 and d[x][y-1] == -1 and (a[x][y-1]&mask)==a[x][y-1])
			d[x][y-1] = d[x][y] + 1, dq.pb({x,y-1});
	}

	if(d[xs][ys] != -1){
		//ans = min(ans,d[xs][ys]);
//		cerr << d[xs][ys] << '\n';
		 if(d[xs][ys] > ans)
		 	return;
		 string x = dfs(xs,ys);
		 if(d[xs][ys] < ans)
		 	ans = d[xs][ys], s = x;
		 else if(d[xs][ys] == ans and x < s)
		 	s = x;
	}
}
main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> n >> m >> k;
	rep(i,1,n){
		rep(j,1,m){
			cin >> ch[i][j];
			if(ch[i][j] != 'T' and ch[i][j] != 'S')
				a[i][j] = (1 << (ch[i][j]-'a'));
			else if(ch[i][j] == 'S')
				xs = i, ys = j;
		    else ye = j , xe = i;
		}
	}
//	check(6);
//	return 0;
	for(int mask = 1; mask < (1 << 26); mask++){
		if(__builtin_popcount(mask) == k){
			check(mask);		
		}
	}
	if(ans == 1e9)
		ret(-1);

	cout << s;
}
