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 <cstdio>
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <sstream>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <bitset>       //UWAGA - w czasie kompilacji musi byc znany rozmiar wektora - nie mozna go zmienic
#include <cassert>
#include <iomanip>      //do setprecision
#include <ctime>
#include <complex>
using namespace std;
#define PB push_back
#define LL long long
#define ULL unsigned LL
#define LD long double

#define MR 210
#define MS 60

int x[4] = {0, 0, -1, 1};
int y[4] = {1, -1, 0, 0};

int n, m, k;

inline bool ok(int i, int w, int k)
{
    return x[i]+w < n && x[i]+w >= 0 && y[i]+k < m && y[i]+k >= 0;
}

int allowed[MR], cost[MS][MS];

char t[MS][MS];

pair < int, int > start, end, prev[MS][MS];

bool cmp(const pair < int, int > &p1, const pair < int, int > &p2)
{
    return t[p1.first][p1.second] < t[p2.first][p2.second];
}

queue < pair < int, int > > q;

string path(2501, 'z'), inf(2501, 'z');

pair < int, int > tab[MR];  //tablica pomocnicza

inline void bfs(int cur)
{
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            cost[i][j] = -1;    
    q.push(start);
    cost[start.first][start.second] = 0;
    vector < char > letters;
    for(int i = 'A'; i <= 'z'; i++)
        if(allowed[i] == cur)
            letters.PB(i);
    while(!q.empty())
    {
        pair < int, int > akt = q.front(); q.pop();     
        for(int j = 0; j < (int)letters.size(); j++)
            for(int i = 0; i < 4; i++)
                if(ok(i, akt.first, akt.second) && t[akt.first+x[i]][akt.second+y[i]] == letters[j])
                {
                    if(cost[akt.first+x[i]][akt.second+y[i]] == -1)
                    {
                        pair < int, int > add(akt.first+x[i], akt.second+y[i]);             
                        cost[add.first][add.second] = cost[akt.first][akt.second] + 1;
                        prev[add.first][add.second] = akt;
                        q.push(add);
                    }               
                }
    }
}//bfs

inline string reconstruct()
{
    string s = "";
    pair < int, int > akt = end;
    while(akt != start)
    {
        s += t[akt.first][akt.second];
        akt = prev[akt.first][akt.second];      
    }
    reverse(s.begin(), s.end());
    s.erase(s.length()-1, 1);
    return s;
}

int main()
{   
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < n; i++)
        scanf("%s", t[i]);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(t[i][j] == 'S')
            {
                start.first = i;
                start.second = j;
            }
            else if(t[i][j] == 'T')
            {
                end.first = i;
                end.second = j;
            }
    int pom = 0;
    if(k == 1)
        for(char a = 'a'; a <= 'z'; a++)
        {
            allowed[a] = allowed['T'] = ++pom;
            bfs(pom);
            string s;
            if(cost[end.first][end.second] != -1)
                s = reconstruct();
            else
                s = inf;
            if(s.length() < path.length() || (s.length() == path.length() && s < path))
                path = s;
        }
    else if(k == 2)
        for(char a = 'a'; a <= 'z'; a++)
            for(char b = a+1; b <= 'z'; b++)
            {                   
                allowed[a] = allowed[b] = allowed['T'] = ++pom;
                bfs(pom);
                string s;
                if(cost[end.first][end.second] != -1)
                    s = reconstruct();
                else
                    s = inf;
                if(s.length() < path.length() || (s.length() == path.length() && s < path))
                    path = s;
            }
    else if(k == 3)
        for(char a = 'a'; a <= 'z'; a++)
            for(char b = a+1; b <= 'z'; b++)
                for(char c = b+1; c <= 'z'; c++)
                {                   
                    allowed[a] = allowed[b] = allowed[c] = allowed['T'] = ++pom;
                    bfs(pom);
                    string s;
                    if(cost[end.first][end.second] != -1)
                        s = reconstruct();
                    else
                        s = inf;
                    if(s.length() < path.length() || (s.length() == path.length() && s < path))
                        path = s;
                }
    else
        for(char a = 'a'; a <= 'z'; a++)
            for(char b = a+1; b <= 'z'; b++)
                for(char c = b+1; c <= 'z'; c++)
                    for(char d = c+1; d <= 'z'; d++)
                    {                   
                        allowed[a] = allowed[b] = allowed[c] = allowed[d] = allowed['T'] = ++pom;
                        bfs(pom);
                        string s;
                        if(cost[end.first][end.second] != -1)
                            s = reconstruct();
                        else
                            s = inf;
                        if(s.length() < path.length() || (s.length() == path.length() && s < path))
                            path = s;
                    }
    if(path != inf)
        printf("%s\n", path.c_str());
    else
        printf("-1\n");
    return 0;
}