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 <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <cstdlib>
#include <time.h>
#include <cassert>
#include <locale.h>
#include <bitset>
#include <fstream>

using namespace std;

#define ll long long
#define ld long double
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define y0 asdfasdf
#define y1 fdsafdsa

template<typename T>
T abs(T x) {
    return x > 0 ? x : -x;
}

template<typename T>
T sqr(T x) {
    return x * x;
}

template<typename T>
void chmin(T &x, T y) {
    x = min(x, y);
}

template<typename T>
void chmax(T &x, T y) {
    x = max(x, y);
}

const int maxn = 55;
const int inf = 1e9;

char a[maxn][maxn];
int dist[maxn][maxn];

const int dirs[4][2] = {
        1, 0,
        0, 1,
        -1, 0,
        0, -1,
};

int n, m, k;

auto valid = [&](int x, int y) {
    return 0 <= x && x < n && 0 <= y && y < m;
};

int sx, sy, tx, ty;

string solve(int allowed) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            dist[i][j] = inf;
        }
    }
    dist[tx][ty] = 0;
    queue<pair<int,int>> q;
    q.push(mp(tx, ty));
    while (!q.empty()) {
        int x = q.front().fst;
        int y = q.front().snd;
        q.pop();
        for (int i = 0; i < 4; i++) {
            int cx = x + dirs[i][0];
            int cy = y + dirs[i][1];
            if (!valid(cx, cy)) {
                continue;
            }
            if (!isupper(a[cx][cy]) && !(allowed & (1 << (a[cx][cy] - 'a')))) {
                continue;
            }
            if (dist[cx][cy] > dist[x][y] + 1) {
                dist[cx][cy] = dist[x][y] + 1;
                q.push(mp(cx, cy));
            }
        }
    }

    string res;
    if (dist[sx][sy] == inf) {
        res += 'z' + 1;
        return res;
    }

    int x = sx, y = sy;
    while (x != tx || y != ty) {
        pair<char, pair<int,int>> bst('z' + 1, mp(-1, -1));
        for (int i = 0; i < 4; i++) {
            int cx = x + dirs[i][0];
            int cy = y + dirs[i][1];
            if (!valid(cx, cy) || dist[cx][cy] + 1 != dist[x][y]) {
                continue;
            }
            chmin(bst, mp(a[cx][cy], mp(cx, cy)));
        }
        res += bst.fst;
        x = bst.snd.fst;
        y = bst.snd.snd;
    }
    res.pop_back();
    return res;
}


string ans;
int mask;
int cnt;

void rec(int p) {
    if (p == 26) {
        chmin(ans, solve(mask));
        return;
    }

    if (cnt + 1 <= k) {
        mask ^= 1 << p;
        cnt++;
        rec(p + 1);
        cnt--;
        mask ^= 1 << p;
    }

    rec(p + 1);
}

int main()
{

#ifdef LOCAL
    freopen("a.in", "r", stdin);
#endif


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

    for (int i = 0; i < n; i++) {
        scanf("%s\n", a[i]);
        for (int j = 0; j < m; j++) {
            if (a[i][j] == 'S') {
                sx = i, sy = j;
            } else if (a[i][j] == 'T') {
                tx = i, ty = j;
            }
        }
    }

    ans += 'z' + 1;

    rec(0);

    if (ans.size() && ans[0] == 'z' + 1) {
        cout << -1 << endl;
    } else {
        cout << ans << endl;
    }

    return 0;
}


