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<fstream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<cctype>
#include<cassert>
#include<cmath>
#include<ctime>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<iomanip>
#include<utility>
using namespace std;
#define mp make_pair
#define pb push_back
#define X first
#define Y second
#define rg register
#define il inline
#define lch(x) ((x)<<1)
#define rch(x) ((x)<<1^1)
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#define rep0(i,n) for(register int i=0;i<(n);++i)
#define per0(i,n) for(register int i=(n)-1;i>=0;--i)
#define rep(i,st,ed) for(register int i=(st);i<=(ed);++i)
#define per(i,ed,st) for(register int i=(ed);i>=(st);--i)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef double dbl;
typedef long double ldb;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<typename T> il T qmin(const T &a,const T &b){return a<b?a:b;}
template<typename T> il T qmax(const T &a,const T &b){return a>b?a:b;}
template<typename T> il void getmin(T &a,const T &b){if(a>b) a=b;}
template<typename T> il void getmax(T &a,const T &b){if(a<b) a=b;}
il void fileio(string s){
	freopen((s+".in").c_str(),"r",stdin);
	freopen((s+".out").c_str(),"w",stdout);
}

const int inf=(int)1e9+7;
const ll linf=(ll)1e17+7;
const int N=55;
const int go[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

namespace calc{
	const int N=2e6+7,M=4e6+7;
	
	int v[M],nxt[M];
	int head[N],dis[N],tp[N];
	bool used[N],can[N],isend[N];
	int node,edge;
	
	il void addedge(int x,int y){
		v[++edge]=y,nxt[edge]=head[x],head[x]=edge;
	}
	
	il void dfs(int u,int rest){
		if(used[u]) return;
		used[u]=true;
		if(!rest){
			if(isend[u]) can[u]=true;
			return;
		}
		for(int e=head[u];e;e=nxt[e]){
			if(dis[u]+1!=dis[v[e]]) continue;
			dfs(v[e],rest-1);
			if(can[v[e]]){
				can[u]=true;
				break;
			}
		}
	}
	
	il void getdis(int from){
		static int q[N],st,ed;
		memset(dis,0,sizeof(dis));
		st=0,ed=1,q[ed]=from,dis[from]=0;
		while(st<ed){
			int u=q[++st];
			for(int e=head[u];e;e=nxt[e]){
				if(v[e]!=from&&!dis[v[e]]){
					dis[v[e]]=dis[u]+1;
					q[++ed]=v[e];
				}
			}
		}
	}
	
	il void getans(int u){
	//	cerr<<u<<endl;
		if(isend[u]) return;
		if(~tp[u]) putchar('a'+tp[u]);
		int to=0;
		for(int e=head[u];e;e=nxt[e]){
			if(dis[v[e]]!=dis[u]+1||!can[v[e]]) continue;
			if(!to||tp[v[e]]<tp[to]) to=v[e];
		}
		getans(to);
	}
	
	il void solve(int from,int tot){
		dfs(from,tot);
		getans(from);
	}
}

int pos[51][51][2952];
map<int,int> id;
int cnt,col[3000];
char g[55][55];
int tp[55][55];
int n,m,k,sti,stj,edi,edj;

il int getv(int i,int j){
	return ~tp[i][j]?1<<tp[i][j]:0;
}

int main(){
//	fileio("test");
	scanf("%d%d%d",&n,&m,&k);
	rep(i,1,n){
		scanf("%s",g[i]+1);
		rep(j,1,m){
			if(islower(g[i][j])) tp[i][j]=g[i][j]-'a';
			else{
				if(g[i][j]=='S') sti=i,stj=j;
				if(g[i][j]=='T') edi=i,edj=j;
				tp[i][j]=-1;
			}
		}
	}
	rep0(i,1<<26){
		if(__builtin_popcount(i)<k){
			id[i]=++cnt;
			col[cnt]=i;
		}
	}
	rep(i,1,n){
		rep(j,1,m){
			int c=tp[i][j];
			rep(t,1,cnt){
				if(~c&&col[t]>>c&1) continue;
				pos[i][j][t]=++calc::node;
				calc::tp[pos[i][j][t]]=c;
			}
		}
	}
	rep(i,1,n){
		rep(j,1,m){
			rep(t,1,cnt){
				int u=pos[i][j][t];
				if(!u) continue;
				int comb=col[t]|getv(i,j);
				rep0(d,4){
					int ii=i+go[d][0],jj=j+go[d][1];
					if(ii<1||ii>n||jj<1||jj>m) continue;
					int ncomb=comb&~(getv(ii,jj)?getv(ii,jj):-1);
					if(!id[ncomb]||!pos[ii][jj][id[ncomb]]) continue;
					calc::addedge(u,pos[ii][jj][id[ncomb]]);
					
				}
			}
		}
	}
	calc::getdis(pos[sti][stj][1]);
	int d=inf;
	rep(t,1,cnt){
		int u=pos[edi][edj][t];
		if(!u||!calc::dis[u]) continue;
		getmin(d,calc::dis[u]);
	}
	rep(t,1,cnt){
		int u=pos[edi][edj][t];
		if(!u||calc::dis[u]!=d) continue;
		calc::isend[u]=true;
	}
	if(d==inf) return puts("-1"),0;
//	cerr<<"d="<<d<<endl;
	calc::solve(pos[sti][stj][1],d);
	puts("");
	return 0;
} 
