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

/*
 * 83C. Track.cpp
 *
 *  Created on: Aug 11, 2015
 *      Author: Osama Hatem
 */

#include <bits/stdtr1c++.h>
#include <ext/numeric>

using namespace std;

struct state {
    int t, r, c, d;
    char ch;

    const bool operator <(const state &s) const {
        if (d != s.d)
            return d < s.d;
        if (t != s.t)
            return t < s.t;
        if (ch != s.ch)
            return ch < s.ch;
        if (r != s.r)
            return r < s.r;
        return c < s.c;
    }
};

int dr[] = { 1, -1, 0, 0 }, dc[] = { 0, 0, 1, -1 };
int n, m, k, sr, sc, er, ec, mask;
char grid[55][55];
bool vis[55][55];
pair<int, int> par[55][55];

int countOnes(int x) {
    int ret = 0;
    while (x) {
        ret += x & 1;
        x >>= 1;
    }
    return ret;
}

bool valid(int r, int c) {
    if (!(r >= 0 && r < n && c >= 0 && c < m && !vis[r][c]))
        return false;
    if ((r == er && c == ec))
        return true;
    return (1 << (grid[r][c] - 'a')) & mask;
}

string Dijkstra() {
    memset(vis, 0, sizeof vis);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            par[i][j] = make_pair(-1, -1);
    bool done = 0;
    set<state> Q;
    Q.insert(state { 0, sr, sc, 0, 'S' });
    while (Q.size()) {
        int r = Q.begin()->r, c = Q.begin()->c, t = Q.begin()->t, d =
                Q.begin()->d;
        Q.erase(Q.begin());
        if (r == er && c == ec) {
            done = 1;
            break;
        }
        if (vis[r][c])
            continue;
        vis[r][c] = 1;
        for (int i = 0; i < 4; i++) {
            int nr = r + dr[i], nc = c + dc[i];
            if (valid(nr, nc)) {
                Q.insert(state { t * 26 + grid[nr][nc] - 'a', nr, nc, d + 1,
                        grid[nr][nc] });
                if (par[nr][nc] == make_pair(-1, -1))
                    par[nr][nc] = make_pair(r, c);
            }
        }
    }
    if (!done)
        return "ST";
    string ret = "";
    int r = er, c = ec;
    while (r != sr || c != sc) {
        int pr = par[r][c].first, pc = par[r][c].second;
        ret += grid[pr][pc];
        r = pr, c = pc;
    }
    reverse(ret.begin(), ret.end());
    ret.erase(ret.begin());
    return ret;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
//  freopen("out.out", "w", stdout);
#endif

    cin >> n >> m >> k;
    for (int i = 0; i < n; i++) {
        cin >> grid[i];
        for (int j = 0; j < m; j++)
            if (grid[i][j] == 'S')
                sr = i, sc = j;
            else if (grid[i][j] == 'T')
                er = i, ec = j;
    }
    string ans = "ST";
    for (mask = 0; mask < 1 << 26; mask++) {
        if (countOnes(mask) != k)
            continue;
        string temp = Dijkstra();
        if (temp == "ST")
            continue;
        if (ans == "ST" || temp.size() < ans.size()
                || (temp.size() == ans.size() && temp < ans))
            ans = temp;
    }
    cout << (ans == "ST" ? "-1" : ans) << endl;
    return 0;
}
