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

#define fr(i, a, x) for(int(i) = a; i <= x; ++i)
#define rfr(i, a, x) for(int(i) = a; i >= x; --i)
#define dist(x1, y1, x2, y2) sqrt( (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) )
#define all(a) a.begin(), a.end()
#define Min(a, b) (a < b) ? a : b
#define Max(a, b) (a > b) ? a : b
#define pb push_back

#include <iostream>
//#include <fstream>
#include <string>
#include <map>
#include <ctime>
#include <cstdlib>
#include <set>
#include <cmath>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
//ifstream cin("input.txt");
//ofstream cout("output.txt");

typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector< vi > vvi;
typedef vector< vb > vvb;
typedef long long ll;
typedef unsigned long long ull;

const int inf = 1e9;

int n, m, k, ei, ej, bi, bj;
vector< vector<char> > b;
vector< vector<string> > g(50, vector<string>(50));
vvi c(50, vi(50));
vb can(256, false), a(256, false);
string best;
ll mn = inf;

int bfs()
{
    queue< pair<int, int> > q;
    q.push(pair<int, int>(bi, bj));
    fr(i, 0, n)
        c[i].assign(m, inf);
    int x, y;
    c[bi][bj] = 0;
    string st;
    while( !q.empty() )
    {
        x = q.front().first;
        y = q.front().second;
        q.pop();
        if (x == ei && y == ej) break;
        if (x + 1 < n && a[b[x + 1][y]])
        {
            if (c[x + 1][y] == c[x][y] + 1)
            {
                st = g[x][y] + b[x + 1][y];
                if (st < g[x + 1][y])
                    g[x + 1][y] = st;
            }
            else if (c[x + 1][y] > c[x][y] + 1)
            {
                g[x + 1][y] = g[x][y] + b[x + 1][y];
                c[x + 1][y] = c[x][y] + 1;
                q.push(pair<int, int>(x + 1, y));
            }
        }
        if (x - 1 >= 0 && a[b[x - 1][y]])
        {
            if (c[x - 1][y] == c[x][y] + 1)
            {
                st = g[x][y] + b[x - 1][y];
                if (st < g[x - 1][y])
                    g[x - 1][y] = st;
            }
            else if (c[x - 1][y] > c[x][y] + 1)
            {
                g[x - 1][y] = g[x][y] + b[x - 1][y];
                c[x - 1][y] = c[x][y] + 1;
                q.push(pair<int, int>(x - 1, y));
            }
        }
        if (y - 1 >= 0 && a[b[x][y - 1]])
        {
            if (c[x][y - 1] == c[x][y] + 1)
            {
                st = g[x][y] + b[x][y - 1];
                if (st < g[x][y - 1])
                    g[x][y - 1] = st;
            }
            else if (c[x][y - 1] > c[x][y] + 1)
            {
                g[x][y - 1] = g[x][y] + b[x][y - 1];
                c[x][y - 1] = c[x][y] + 1;
                q.push(pair<int, int>(x, y - 1));
            }
        }
        if (y + 1 < m && a[b[x][y + 1]])
        {
            if (c[x][y + 1] == c[x][y] + 1)
            {
                st = g[x][y] + b[x][y + 1];
                if (st < g[x][y + 1])
                    g[x][y + 1] = st;
            }
            else if (c[x][y + 1] > c[x][y] + 1)
            {
                g[x][y + 1] = g[x][y] + b[x][y + 1];
                c[x][y + 1] = c[x][y] + 1;
                q.push(pair<int, int>(x, y + 1));
            }
        }
    }
    return c[ei][ej];
}

void f(int h, char qw)
{
    if (h == k + 1)
    {
        ll w = bfs();
        if (w < mn)
        {
            mn = w;
            best = g[ei][ej];
        }
        if (w == mn && best > g[ei][ej])
            best = g[ei][ej];
        return;
    }
    else 
    {
        bool t = false;
        fr(i, qw, 'z')
            if (can[i])
            {
                a[i] = true;
                f(h + 1, i + 1);
                a[i] = false;
                t = true;
            }
        if (!t)
        {
            ll w = bfs();
            if (w < mn)
            {
                mn = w;
                best = g[ei][ej];
            }
            if (w == mn && best > g[ei][ej])
                best = g[ei][ej];
            return;
        }
    }
    return;
}

int main()
{
    cin >> n >> m >> k;
    b.resize(n);
    fr(i, 0, n - 1)
    {
        b[i].resize(m);
        fr(j, 0, m - 1)
        {
            cin >> b[i][j];
            if (b[i][j] == 'S')
            {
                bi = i;
                bj = j;
            }
            if (b[i][j] == 'T')
            {
                ei = i;
                ej = j;
            }
            can[b[i][j]] = true;
        }   
    }
    char t = 'z' + 1;
    best += t;
    a['T'] = a['S'] = true;
    if (bi - 1  >= 0)
    {
        a[b[bi - 1][bj]] = true;
        f(2, 'a');
        a[b[bi - 1][bj]] = true;
    }
    if (bi + 1 < n)
    {
        a[b[bi + 1][bj]] = true;
        f(2, 'a');
        a[b[bi + 1][bj]] = true;
    }
    if (bj - 1  >= 0)
    {
        a[b[bi][bj - 1]] = true;
        f(2, 'a');
        a[b[bi][bj - 1]] = true;
    }
    if (bj + 1 < n)
    {
        a[b[bi][bj + 1]] = true;
        f(2, 'a');
        a[b[bi][bj + 1]] = true;
    }
    if (mn == inf) 
    {
        cout << -1 << '\n';
        return 0;
    }
    if (best.size() > 1)
        fr(i, 0, best.size() - 2)
            cout << best[i];
    cout << '\n';
}