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.util.InputMismatchException;
import java.io.*;
import java.util.Arrays;

/**
 * Generated by Contest helper plug-in
 * Actual solution is at the bottom
 */
public class Main {
    public static void main(String[] args) {
        InputReader in = new StreamInputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        run(in, out);
    }

    public static void run(InputReader in, PrintWriter out) {
        Solver solver = new taskC();
        solver.solve(1, in, out);
        Exit.exit(in, out);
    }
}

abstract class InputReader {
    private boolean finished = false;

    public abstract int read();

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }
    public String readString() {
        return readLine();
    }


    public String readLine() {
        StringBuffer res = new StringBuffer();
        int c = read();
        while (c != '\n' && c != -1) {
            if (c != '\r')
                res.appendCodePoint(c);
            c = read();
        }
        return res.toString();
    }
    public String nextToken() {
        int c = read();
        while (isSpaceChar(c)) {
            c = read();
        }
        StringBuffer res = new StringBuffer();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }
    private boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public void setFinished(boolean finished) {
        this.finished = finished;
    }

    public abstract void close();
}

class StreamInputReader extends InputReader {
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar, numChars;

    public StreamInputReader(InputStream stream) {
        this.stream = stream;
        curChar = 0;
    }

    public int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    @Override
    public void close() {
        try {
            stream.close();
        } catch (IOException ignored) {
        }
    }
}

class Exit {
    private Exit() {
    }

    public static void exit(InputReader in, PrintWriter out) {
        in.setFinished(true);
        in.close();
        out.close();
    }
}

interface Solver {
    public void solve(int testNumber, InputReader in, PrintWriter out);
}

class taskC implements Solver {
    int n, m;
    char s[][];
    int dist[][];
    int prev[][][];
    int ans = -1;
    String ansString = "";
    boolean can[];

    int startX , startY;
    int finishX, finishY;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        n = in.nextInt();
        m = in.nextInt();
        int k = in.nextInt();
        s = new char[n][];
        for (int i =0 ; i < n; ++i) {
            s[i] = in.nextToken().toCharArray();
        }
        for (int i =0 ; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (s[i][j] == 'S') {
                    startX = j;
                    startY = i;
                }
                if (s[i][j] == 'T') {
                    finishX = j;
                    finishY = i;
                }
            }
        }
        can = new boolean[26];

        qx = new int[n * m];
        qy = new int[n * m];
        dist = new int[n][m];
        prev = new int[n][m][2];
        brute(0, k);

        if (ans == -1) {
            out.print(ans);
            return;
        }
        for (int i = 0; i < ansString.length() - 1; ++i) {
            out.print(ansString.charAt(i));
        }

    }
    int qx[];
    int qy[];
    int dx[] = new int[] {0, 0, -1, 1};
    int dy[] = new int[] {-1, 1, 0, 0};
    boolean valid(int x, int y) {
        return x >= 0 && y >= 0 && x < m && y < n && (s[y][x] == 'T' || s[y][x] == 'S' || can[s[y][x] - 'a']) && dist[y][x] == -1;
    }
    void brute(int pos, int k) {
        if (pos == 26) {

            for (int [] d : dist) Arrays.fill(d, -1);
            int head =0;
            dist[startY][startX] = 0;
            qx[0] = startX;
            qy[0] = startY;
            head ++;
            for (int i = 0; i < head; ++i) {
                int x = qx[i];
                int y = qy[i];
                int cnt = 0;
                for (int u = 0; u < 4; ++u) {
                    int nx = x + dx[u];
                    int ny = y + dy[u];
                    if (valid(nx, ny)) {
                        ++ cnt;
                    }
                }
                int waysX[] = new int[cnt];
                int waysY[] = new int[cnt];
                int waysC[] = new int[cnt];
                cnt = 0;
                for (int u = 0; u < 4; ++u) {
                    int nx = x + dx[u];
                    int ny = y + dy[u];
                    if (valid(nx, ny)) {
                        waysY[cnt] = ny;
                        waysX[cnt] = nx;
                        waysC[cnt] = s[ny][nx];
                        ++ cnt;
                    }
                }
                for (int u = 0; u < cnt; ++u) {
                    for (int v = 0; v < cnt - 1; ++v) {
                        if (waysC[v] > waysC[v + 1]) {
                            int tmp = waysX[v]; waysX[v] = waysX[v + 1]; waysX[v + 1] = tmp;
                            tmp = waysY[v]; waysY[v] = waysY[v + 1]; waysY[v + 1] = tmp;
                            tmp = waysC[v]; waysC[v] = waysC[v + 1]; waysC[v + 1] = tmp;
                        }
                    }
                }
                for (int u = 0; u < cnt; ++u) {
                    if (valid(waysX[u], waysY[u])) {
                        qx[head] = waysX[u];
                        qy[head] = waysY[u];
                        dist[waysY[u]][waysX[u]] = dist[y][x] + 1;
                        prev[waysY[u]][waysX[u]][0] = x;
                        prev[waysY[u]][waysX[u]][1] = y;
                        head ++ ;
                    }
                }
            }
            /*System.out.println(Arrays.toString(can));
            for (int i = 0; i < dist.length; ++i) {
                System.out.println(Arrays.toString(dist[i]));
            }
            System.out.println("\n\n");*/
            if (dist[finishY][finishX] != -1) {
                if (ans == -1 || ans >= dist[finishY][finishX]) {
                    int cx = finishX;
                    int cy = finishY;
                    StringBuffer sbf = new StringBuffer() ;
                    while (cx != startX || cy != startY) {
                        sbf.append(s[cy][cx]);
                        int tx = prev[cy][cx][0];
                        int ty = prev[cy][cx][1];
                        cx = tx;
                        cy = ty;
                    }
                    sbf.reverse();
                    String a = sbf.toString();
                    if (ans == -1 || ans > dist[finishY][finishX] || ans == dist[finishY][finishX] &&  a.compareTo(ansString) < 0) {
                        ans = dist[finishY][finishX] ;
                        ansString = a;
                    }
                }
            }
        } else {
            if (k > 0) {
                can[pos] = true;
                brute(pos + 1, k - 1);
            }
            can[pos] = false;
            brute(pos + 1, k);
        }
    }

}
