Problem: 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

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class C {

    public static void main(String[] args) throws IOException {
        new C().solve();
    }

    Pair[] go = {newPair(-1, 0), newPair(1, 0), newPair(0, -1), newPair(0, 1)};

    StringTokenizer st = new StringTokenizer("");
    BufferedReader br;
    int n;
    int m;
    char[][] a;
    Pair s;
    Pair t;
    Pair[][][] vandsg;
    String bestPath;
    Map<String, String> done = new HashMap<String, String>();

    private void solve() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        n = nextInt();
        m = nextInt();
        int k = nextInt();
        a = new char[n][m];
        for (int i = 0; i < n; i++) {
            String s = nextToken();
            for (int j = 0; j < s.length(); j++) {
                a[i][j] = s.charAt(j);
            }
        }
        br.close();
        Set<Character> allTypes = new HashSet<Character>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                char type = a[i][j];
                switch (type) {
                    case 'S':
                        s = newPair(i, j);
                        break;
                    case 'T':
                        t = newPair(i, j);
                        break;
                    default:
                        allTypes.add(type);
                        break;
                }
            }
        }
        vandsg = new Pair[n][m][];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                Pair p = newPair(i, j);
                List<Pair> pairs = new ArrayList<Pair>(go.length);
                for (Pair v : go) {
                    Pair n = add(p, v);
                    if (check(n)) {
                        pairs.add(n);
                    }
                }
                Collections.sort(pairs, new Comparator<Pair>() {
                    public int compare(Pair p1, Pair p2) {
                        return charAt(p1) - charAt(p2);
                    }
                });
                Pair[] r = pairs.toArray(new Pair[pairs.size()]);
                vandsg[i][j] = r;
            }
        }
        for (Pair sGo : go) {
            Pair p1 = add(s, sGo);
            for (Pair tGo : go) {
                Pair t1 = add(t, tGo);
                if (check(p1) && check(t1)) {
                    List<Character> rest = new ArrayList<Character>(allTypes);
                    rest.remove(Character.valueOf(charAt(p1)));
                    rest.remove(Character.valueOf(charAt(t1)));
                    boolean[] allowedTypes = new boolean[256];
                    allowedTypes[charAt(p1)] = true;
                    allowedTypes[charAt(t1)] = true;
                    allowedTypes['S'] = true;
                    allowedTypes['T'] = true;
                    rec(allowedTypes, rest, 0, k - (allTypes.size() - rest.size()));
                }
            }
        }
        PrintWriter pw = new PrintWriter(System.out);
        if (bestPath == null) {
            pw.println(-1);
        } else {
            pw.println(bestPath);
        }
//        if ("aadbaaabbbdbabbabbabbbbaaaaadaabaabbdbaababbbd".equals(bestPath)) {
//            for (Map.Entry<String, String> stringStringEntry : done.entrySet()) {
//                pw.println(stringStringEntry.getKey() + ": " + stringStringEntry.getValue());
//            }
//        }
        pw.close();
    }

    void rec(boolean[] allowedTypes, List<Character> rest, final int fromIndex, final int restCount) {
        if (restCount < 0) {
            return;
        }
        if (restCount == 0 || rest.size() == fromIndex) {
            doIt(allowedTypes);
        }
        for (int i = fromIndex; i < rest.size(); i++) {
            char newType = rest.get(i);
            if (allowedTypes[newType]) {
                throw new RuntimeException("BOTVA");
            }
            allowedTypes[newType] = true;
            rec(allowedTypes, rest, i + 1, restCount - 1);
            allowedTypes[newType] = false;
        }
    }

    void doIt(boolean[] allowedTypes) {
        StringBuilder allowedSb = new StringBuilder();
        for (int i = 0; i < allowedTypes.length; i++) {
            if (allowedTypes[i]) {
                allowedSb.append((char) i);
            }
        }
        final String allowedStr = allowedSb.toString();
        if (done.containsKey(allowedStr)) {
            return;
        }
//        solve1(allowedTypes);

        Pair[][] prev = new Pair[n][m];
        List<Pair>[] pairsByChar = new List[256];
        for (int i = 'a'; i <= 'z'; i++) {
            pairsByChar[i] = new ArrayList<Pair>();
        }
        List<Pair> current = new ArrayList<Pair>();
        current.add(t);
        while (prev[s.i][s.j] == null && current.size() > 0) {
            for (Pair p : current) {
                Pair[] goes = vandsg[p.i][p.j];
                for (Pair n : goes) {
                    char nch = charAt(n);
                    if (allowedTypes[nch] && prev[n.i][n.j] == null) {
                        prev[n.i][n.j] = p;
                        if ('a' <= nch && nch <= 'z') {
                            pairsByChar[nch].add(n);
                        }
                    }
                }
            }
            current.clear();
            for (char ch = 'a'; ch <= 'z'; ch++) {
                current.addAll(pairsByChar[ch]);
                pairsByChar[ch].clear();
            }
        }
        int len = solve1(allowedTypes);
        if (prev[s.i][s.j] == null) {
            if (len != -1) {
                throw new RuntimeException("false positive");
            }
            done.put(allowedStr, "<no path>");
            return;
        }
        StringBuilder sb = new StringBuilder();
        Pair p = s;
        while (p.i != t.i || p.j != t.j) {
            if (p != s) {
                sb.append(charAt(p));
            }
            p = prev[p.i][p.j];
        }
        String path = sb.toString();
        done.put(allowedStr, path);
        if (path.length() != len) {
            throw new RuntimeException("length are unequal");
        }
        if (bestPath == null) {
            bestPath = path;
        } else if (path.length() < bestPath.length()) {
            bestPath = path;
        } else if (path.length() == bestPath.length() && path.compareTo(bestPath) < 0) {
            bestPath = path;
        }
    }

    int solve1(boolean[] allowedTypes) {
        int[][] a = new int[n][m];
        for (int[] ints : a) {
            Arrays.fill(ints, -1);
        }
        a[s.i][s.j] = 0;
        Queue<Pair> q = new LinkedList<Pair>();
        q.add(s);
        while (a[t.i][t.j] == -1 && q.size() > 0) {
            Pair p = q.poll();
            int d = a[p.i][p.j];
            Pair[] goes = vandsg[p.i][p.j];
            for (Pair n : goes) {
                char nch = charAt(n);
                if (a[n.i][n.j] == -1 && allowedTypes[nch]) {
                    a[n.i][n.j] = d + 1;
                    q.add(n);
                }
            }
        }
        if (a[t.i][t.j] == -1) {
            return -1;
        } else {
            return a[t.i][t.j] - 1;
        }
    }

    char charAt(Pair p) {
        return a[p.i][p.j];
    }

    Pair add(Pair a, Pair b) {
        return newPair(a.i + b.i, a.j + b.j);
    }

    boolean check(Pair p) {
        return 0 <= p.i && p.i < n && 0 <= p.j && p.j < m;
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    String nextToken() throws IOException {
        while (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    Pair newPair(int i, int j) {
        return new Pair(i, j);
    }

    static class Pair {
        final int i, j;

        Pair(int i, int j) {
            this.i = i;
            this.j = j;
        }

        @Override
        public String toString() {
            return "(" + i + ", " + j + ")";
        }
    }
}