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
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.regex.*;
import static java.lang.Math.*;
public class C {
    public static final int INF = 1<<28;
    public static final int[] X = {1, 0, -1, 0};
    public static final int[] Y = {0, 1, 0, -1};
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String line;
    StringTokenizer st;
    StringBuilder buf = new StringBuilder();
    public C() throws Exception {
        line = br.readLine();
        st = new StringTokenizer(line);
        int m = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        
        Set<Character> letters = new HashSet();
        char[][] map = new char[m][n];
        int sx = -1;
        int sy = -1;
        int tx = -1;
        int ty = -1;
        for (int i=0;i<m;i++) {
            map[i] = br.readLine().toCharArray();
            for (int j=0;j<n;j++) {
                if (map[i][j]=='S') {
                    sx = i;
                    sy = j;
                } else if(map[i][j]=='T') {
                    tx = i;
                    ty = j;
                } else 
                    letters.add(map[i][j]);
            }
        }

        if (k>letters.size())
            k = letters.size();

        char[] set = new char[letters.size()];
        {
            int i = 0;
            for (char c:letters) {
                set[i] = c;
                i++;
            }
        }
        Arrays.sort(set);

        int best = INF;
        String ans = null;

        int[] v = new int[k+1];
        v[0] = -1;
        int pos = 0;
        while (true) {
            if (pos==k) {
                boolean[] used = new boolean[128];
                used['S'] = true;
                for (int i=1;i<=k;i++) {
                    used[set[v[i]]] = true;
                }

                LinkedList<Integer> queue = new LinkedList<Integer>();
                int[][] dis = new int[m][n];
                for (int i=0;i<m;i++) {
                    Arrays.fill(dis[i], INF);
                }
                
                queue.offer(tx);
                queue.offer(ty);
                dis[tx][ty] = 0;
                
                while(!queue.isEmpty()){
                    int cx = queue.poll();
                    int cy = queue.poll();
                
                    for (int i=0;i<4;i++) {
                        int nx = cx+X[i];
                        int ny = cy+Y[i];
                        if (nx<0||nx>=m||ny<0||ny>=n) continue;
                        if (dis[nx][ny]!=INF) continue;
                        if (!used[map[nx][ny]]) continue;
                        dis[nx][ny] = dis[cx][cy]+1;
                        queue.offer(nx);
                        queue.offer(ny);
                    }
                }


                int d = dis[sx][sy];
                // debug(best, d);
                if (d!=INF&&best>=d-1) {

                    queue = new LinkedList<Integer>();
                    queue.offer(sx);
                    queue.offer(sy);
                    
                    StringBuilder ansBuf = new StringBuilder();
                    int t = d;
                    while(!queue.isEmpty()){
                        t--;
                        int minC = INF;
                        for (int j=0;j<queue.size();j+=2) {
                            int cx = queue.get(j);
                            int cy = queue.get(j+1);
                            for (int p=0;p<4;p++) {
                                int nx = cx+X[p];
                                int ny = cy+Y[p];
                                if (nx<0||nx>=m||ny<0||ny>=n) continue;
                                if (dis[nx][ny]!=t) continue;
                                minC = min(minC, map[nx][ny]);
                            } 
                        } 

                        for (int bfs=queue.size();bfs>0;bfs-=2) {
                            int cx = queue.poll();
                            int cy = queue.poll();
                            for (int p=0;p<4;p++) {
                                int nx = cx+X[p];
                                int ny = cy+Y[p];
                                if (nx<0||nx>=m||ny<0||ny>=n) continue;
                                if (dis[nx][ny]!=t) continue;
                                if (map[nx][ny]==minC) {
                                    queue.offer(nx);
                                    queue.offer(ny);
                                    dis[nx][ny] = -1;
                                }
                            } 
                        }
                        if (minC=='T') break;
                        ansBuf.append((char)minC);
                    }


                    String str = ansBuf.toString();
                    // debug(1, str);
                    if (best>d-1||ans.compareTo(str)>0)
                        ans = str;
                    best = d-1;
                } 
            }
            if (v[pos]<set.length-1) {
                if (pos==k) {
                    v[pos]++;
                } else {
                    v[pos+1] = v[pos]+1;
                    pos++;
                }
            } else if (pos!=0) {
                v[--pos]++;
            }
            if (pos==0) break;
        }

        if (ans==null)
            ans = "-1";
        buf.append(ans).append('\n'); 
        System.out.print(buf);
    }

    public static void main(String[] args) throws Exception { // {{{
        new C();
    } // }}}
    public static void debug(Object... arr) { // {{{
        System.err.println(Arrays.deepToString(arr));
    } // }}}
}
