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

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

/**
 * Author -
 * User: kansal
 * Date: 5/13/11
 * Time: 10:07 PM
 */
public class Track {
    public static void main(String[] args) {
        reader = new BufferedReader(new InputStreamReader(System.in));

        int n = nextInt(), m = nextInt(), k = nextInt();

        String[] B = new String[n];
        for(int i = 0; i < n; ++i) B[i] = nextLine();

        int totalLetters = 0;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                switch (B[i].charAt(j)) {
                    case 'S':
                        startX = i; startY = j;
                        break;
                    case 'T':
                        endX = i; endY = j;
                        break;
                    default:
                        if (!isThere[B[i].charAt(j)-'a']) ++totalLetters;
                        isThere[B[i].charAt(j)-'a'] = true;
                }
            }
        }

        k = Math.min(k, totalLetters);
        doit(0, k, B);

        if (minD == 1234567890) {
            System.out.println(-1);
        }
        else {
            System.out.println(Res);
        }
    }

    static String Res = "zzz";
    static int minD = 1234567890;
    static char[] use = new char[4];
    static boolean[] isThere = new boolean[26];
    static int startX = -1, startY = -1;
    static int endX = -1, endY = -1;

    static void doit(int z, int k, String[] B) {
        if (z < k) {
            for(char c = 'a'; c <= 'z'; ++c) {
                if (isThere[c-'a']) {
                    if (z > 0 && use[z-1] >= c) continue;
                    use[z] = c;
                    doit(z+1, k, B);
                }
            }
            return;
        }

        bfs(k, B);
    }

    static void bfs(int K, String[] B) {
        int n = B.length, m = B[0].length();

        int[][] d = new int[n][m];
        for(int[] dd: d) Arrays.fill(dd, 1234567890);
        d[startX][startY] = 0;

        int[] q = new int[n*m+100];
        int qh = 0, qt = 0;
        q[qt++] = startX * m + startY;

        int[][] prev = new int[n][m];
        while (qh < qt) {
            int cur = q[qh++];
            int x = cur / m, y = cur % m;

            if (x == endX && y == endY) {
                break;
            }

            final char[] cV = new char[4];
            int[] iV = new int[4];
            int z = 0;
            for(int dx = -1; dx <= 1; ++dx) {
                for(int dy = -1; dy <= 1; ++dy) {
                    if (dx == 0 ^ dy == 0) {
                        int nx = x + dx, ny = y + dy;
                        if (nx < 0 || nx == n || ny < 0 || ny == m) {
                            continue;
                        }

                        char c = B[nx].charAt(ny);
                        boolean canUse = c == 'T';
                        for(int i = 0; i < K; ++i) if (use[i] == c) canUse = true;
                        if (!canUse) continue;

                        if (d[nx][ny] <= d[x][y] + 1) {
                            continue;
                        }

                        cV[z] = c;
                        iV[z++] = nx * m + ny;
                        prev[nx][ny] = x * m + y;
                        d[nx][ny] = d[x][y] + 1;
                    }
                }
            }

            Integer[] order = new Integer[4];
            for(int i = 0; i < 4; ++i) order[i] = i;

            Arrays.sort(order, 0, z, new Comparator<Integer>() {
                public int compare(Integer o1, Integer o2) {
                    return cV[o1] - cV[o2];
                }
            });

            for(int i = 0; i < z; ++i) {
                q[qt++] = iV[order[i]];
            }
        }

        if (d[endX][endY] == 1234567890 || d[endX][endY] > minD) {
            return;
        }

        String res = "";

        int X = endX, Y = endY;
        int p = prev[X][Y];
        X = p / m; Y = p % m;

        while (X != startX || Y != startY) {
            res = B[X].charAt(Y) + res;
            p = prev[X][Y];
            X = p / m; Y = p % m;
        }

        if (d[endX][endY] < minD || res.compareTo(Res) < 0) {
            minD = d[endX][endY];
            Res = res;
        }
    }


    public static BufferedReader reader;

    public static StringTokenizer tokenizer = null;

    static String nextToken() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    static public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    static public String next() {
        return nextToken();
    }

    static public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
}
