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 <string>
#include <string.h>
#include <queue>
#include <math.h>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <bitset>
#include <list>
#include <ctype.h>
#include <cassert>
#include <stack>
#include <fstream>
#include <unordered_map>
#include <unordered_set>
#include <ctime>
#include <functional>
#include <ctime>
#include <limits>
#include <complex>

using namespace std;

#define snd second
#define fst first
#define mp make_pair
#define ll long long
#define ull unsigned long long
#define ld double
#define pb push_back
#define left _left
#define right _right

const ld pi = acos(-1.0l);

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);
}

template<typename U, typename V>
ostream &operator<<(ostream &s, const pair<U, V> &x) {
    s << "(" << x.fst << ", " << x.snd << ")";
    return s;
}

template<typename U>
ostream &operator<<(ostream &s, const vector<U> &x) {
    s << "[";
    bool was = false;
    for (auto it : x) {
        if (was) {
            s << ", ";
        }
        was = true;
        s << it;
    }
    s << "]";
    return s;
}

template<typename U>
ostream &operator<<(ostream &s, const set<U> &x) {
    s << "{";
    bool was = false;
    for (auto it : x) {
        if (was) {
            s << ", ";
        }
        was = true;
        s << it;
    }
    s << "}";
    return s;
}

template<int sz>
ostream operator<<(ostream &s, const bitset<sz> &x) {
    for (int i = 0; i < sz; i++) {
        s << x[i];
    }
    return s;
}


//-----------------------------------------------------------------------------


vector<string> a;
int sx, sy, tx, ty;

bool valid(int x, int y) {
    return 0 <= x && x < a.size() && 0 <= y && y < a[0].size();
}

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

string solve(int allowed) {
    const int inf = 1e9;
    int n = a.size(), m = a[0].size();

    vector<vector<int>> dist(n, vector<int>(m, inf));
    dist[sx][sy] = 0;

    queue<pair<int, int>> q;
    q.push(mp(sx, sy));

    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 (islower(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));
            }
        }
    }

    if (dist[tx][ty] == inf) {
        string res(50 * 50 + 1, 'z' + 1);
        return res;
    }

    string res;
    vector<pair<int, int>> c({mp(sx, sy)});
    for (int iter = 0; iter < dist[tx][ty]; iter++) {
        char bst = 0;
        vector<pair<int, int>> nc;
        for (auto p : c) {
            for (int i = 0; i < 4; i++) {
                int cx = p.fst + dirs[i][0];
                int cy = p.snd + dirs[i][1];
                if (!valid(cx, cy) || dist[cx][cy] != iter + 1) {
                    continue;
                }
                if (!bst || bst > a[cx][cy]) {
                    bst = a[cx][cy];
                    nc.clear();
                    nc.pb(mp(cx, cy));
                } else if (bst == a[cx][cy]) {
                    nc.pb(mp(cx, cy));
                }
            }
        }
        res += bst;
        c = nc;
    }

    res.pop_back();
    return res;
}


int main() {
    srand(time(NULL));

    retry:

#ifdef LOCAL
    //gen();
    //cerr << 1 << endl;
    freopen("a.in", "r", stdin);
#else
    //freopen("fract.in", "r", stdin);
    //freopen("fract.out", "w", stdout);
#endif

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

    a.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> 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;
            }
        }
    }

    int cnt = 0, allowed = 0;
    string ans(50 * 50 + 1, 'z' + 1);

    function<void(int)> rec;
    rec = [&](int p) {
        if (p == 26) {
            auto r = solve(allowed);
            if (r.size() < ans.size()) {
                ans = r;
            } else if (r.size() == ans.size()) {
                chmin(ans, r);
            }
            return;
        }

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

        rec(p + 1);
    };

    rec(0);

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

    return 0;
}