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 <algorithm>
#include <cstring>
#include <climits>
#include <cmath>
#include <queue>
#include <cstdio>
#include <functional>
#include <vector>





using namespace std;

typedef pair<int, int> pii;
#define mp make_pair

typedef struct link
{
    pii t;
    link* next;
}link;

char mat[55][55];
int dist[55][55];
bool vis[300];
bool ok[30];
int array[30];
int cnt;
int c;

link g[10000];
link* head[2510];

int n, m, k;
char ans[10000];

pii q[10000];
bool valid[55][55];

bool isOk(pii a)
{
    if(a.first >= 0 && a.first < n && a.second >= 0 && a.second < m)
    return true;
    return false;
}

void add_link(int s, pii t)
{
    g[cnt].t = t;
    g[cnt].next = head[s];
    head[s] = &g[cnt++];
}

void BFS()
{
    pair<int, int> s, t;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            if('S' == mat[i][j])
            {
                s = make_pair(i, j);
            }
            else if('T' == mat[i][j])
            {
                t = make_pair(i, j);
            }
        }
    }

    int is = 0, it = 0;
    q[it++] = t;
    memset(dist, -1, sizeof(dist));
    dist[t.first][t.second] = 0;
    pii u, v;
    while(is < it)
    {
        u = q[is++];
        v = mp(u.first, u.second + 1);
        if(isOk(v) && ('S' == mat[v.first][v.second] || ok[mat[v.first][v.second] - 'a']) && -1 == dist[v.first][v.second])
        {
            q[it++] = v;
            dist[v.first][v.second] = dist[u.first][u.second] + 1;
        }
        v = mp(u.first, u.second - 1);
        if(isOk(v) && ('S' == mat[v.first][v.second] || ok[mat[v.first][v.second] - 'a']) && -1 == dist[v.first][v.second])
        {
            q[it++] = v;
            dist[v.first][v.second] = dist[u.first][u.second] + 1;
        }
        v = mp(u.first + 1, u.second);
        if(isOk(v) && ('S' == mat[v.first][v.second] || ok[mat[v.first][v.second] - 'a']) && -1 == dist[v.first][v.second])
        {
            q[it++] = v;
            dist[v.first][v.second] = dist[u.first][u.second] + 1;
        }
        v = mp(u.first - 1, u.second);
        if(isOk(v) && ('S' == mat[v.first][v.second] || ok[mat[v.first][v.second] - 'a']) && -1 == dist[v.first][v.second])
        {
            q[it++] = v;
            dist[v.first][v.second] = dist[u.first][u.second] + 1;
        }
    }
    if(-1 == dist[s.first][s.second])
    return;
    memset(head, 0, sizeof(head));
    cnt = 0;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            if(-1 != dist[i][j])
            {
                add_link(dist[i][j], mp(i, j));
            }
        }
    }
    int d = dist[s.first][s.second];
    memset(valid, false, sizeof(valid));
    valid[s.first][s.second] = true;
    char tmp[1000000];

    for(int i = d; i >= 2; i--)
    {
        char nMin = 'z';
        for(link* p = head[i]; p; p = p->next)
        {
            if(valid[p->t.first][p->t.second])
            {
                v = mp(p->t.first, p->t.second + 1);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    nMin = min(nMin, mat[v.first][v.second]);
                }
                v = mp(p->t.first, p->t.second - 1);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    nMin = min(nMin, mat[v.first][v.second]);
                }
                v = mp(p->t.first + 1, p->t.second);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    nMin = min(nMin, mat[v.first][v.second]);
                }
                v = mp(p->t.first - 1, p->t.second);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    nMin = min(nMin, mat[v.first][v.second]);
                }
            }
        }
        tmp[d - i] = nMin;
        for(link* p = head[i]; p; p = p->next)
        {
            if(valid[p->t.first][p->t.second])
            {
                v = mp(p->t.first, p->t.second + 1);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    if(nMin == mat[v.first][v.second])
                    valid[v.first][v.second] = true;
                }
                v = mp(p->t.first, p->t.second - 1);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    if(nMin == mat[v.first][v.second])
                    valid[v.first][v.second] = true;
                }
                v = mp(p->t.first + 1, p->t.second);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    if(nMin == mat[v.first][v.second])
                    valid[v.first][v.second] = true;
                }
                v = mp(p->t.first - 1, p->t.second);
                if(isOk(v) && ok[mat[v.first][v.second] - 'a'] && dist[v.first][v.second] + 1 == i)
                {
                    if(nMin == mat[v.first][v.second])
                    valid[v.first][v.second] = true;
                }
            }
        }
    }
    tmp[d] = 0;
    if(-1 == ans[0])
    {
        strcpy(ans, tmp);
    }
    else if(strcmp(ans, tmp) > 0)
    {
        strcpy(ans, tmp);
    }
}







int main()
{
//    freopen("input.txt", "r", stdin);

    scanf("%d %d %d", &n, &m, &k);


    for(int i = 0; i < n; i++)
    {
        scanf("%s", mat[i]);
    }
    if(m * n == 2)
    {
        puts("");
        return 0;
    }

    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            if(mat[i][j] >= 'a' && mat[i][j] <= 'z')
            vis[mat[i][j] - 'a'] = true;
        }
    }
    for(int i = 0; i < 26; i++)
    if(vis[i])
    array[c++] = i;
    ans[0] = -1;

    switch(k)
    {
        case 1:
        {
            for(int i = 0; i < c; i++)
            {
                memset(ok, false, sizeof(ok));
                ok[array[i]] = true;
                BFS();
            }
            break;
        }
        case 2:
        {


            for(int i = 0; i < c; i++)
            {
                for(int j = i + 1; j < c; j++)
                {
                    memset(ok, false, sizeof(ok));
                    ok[array[i]] = ok[array[j]] = true;

                    BFS();
                }
            }
            if(c < 2)
            {
                memset(ok, false, sizeof(ok));
                for(int i = 0; i < c; i++)
                ok[array[i]] = true;
                BFS();
            }
            break;
        }
        case 3:
        {
            for(int i = 0; i < c; i++)
            {
                for(int j = i + 1; j < c; j++)
                {
                    for(int k = j + 1; k < c; k++)
                    {
                        memset(ok, false, sizeof(ok));
                        ok[array[i]] = ok[array[j]] = ok[array[k]] = true;
                        BFS();
                    }
                }
            }
            if(c < 3)
            {
                memset(ok ,false, sizeof(ok));
                for(int i = 0; i < c; i++)
                {
                    ok[array[i]] = true;
                }
                BFS();
            }
            break;
        }
        case 4:
        {
            for(int i = 0; i < c; i++)
            {
                for(int j = i + 1; j < c; j++)
                {
                    for(int k = j + 1; k < c; k++)
                    {
                        for(int l = k + 1; l < c; l++)
                        {
                            memset(ok, false, sizeof(ok));
                            ok[array[i]] = ok[array[j]] = ok[array[k]] = ok[array[l]] = true;
                            BFS();
                        }
                    }
                }
            }
            if(c < 4)
            {
                memset(ok, false, sizeof(ok));
                for(int i = 0; i < c; i++)
                {
                    ok[array[i]] = true;
                }
                BFS();
            }
            break;
        }

    }
    if(ans[0] == -1)
    puts("-1");
    else
    puts(ans);










    return 0;
}
