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 static java.util.Arrays.*;
import static java.lang.Math.*;
import static java.math.BigInteger.*;
import java.util.*;
import java.math.*;
import java.io.*;

public class C implements Runnable
{
    String file = "input";
    
    boolean TEST = false;
    
    void solve() throws IOException
    {
        rows = nextInt(); 
        cols = nextInt(); 
        K = nextInt();
        cs = new char[rows][];
        for(int i = 0; i < rows; i++)
        { 
            cs[i] = next().toCharArray();
            for(int j = 0; j < cols; j++) 
                if(cs[i][j] == 'S')
                {
                    sx = i;
                    sy = j;
                }
                else if(cs[i][j] == 'T')
                {
                    ex = i;
                    ey = j;
                }
        }
        res = null;
        go(0, 0, 0);
        out.println(res == null ? -1 : res);
    }
    char[][] cs;
    int sx, sy, ex, ey;
    int rows, cols, K;

    void go(int pos, int num, int mask)
    {
        if(num > K) return;
        if(pos == 26)
        {
            get(mask);
            return;
        }
        go(pos + 1, num, mask);
        go(pos + 1, num + 1, mask | (1 << pos));
    }
    int[][] d;
    String res = null;
    void get(int mask)
    {
        BFS(ex, ey, mask);
        if(d[sx][sy] >= INF) return;
        if(res != null && d[sx][sy] - 1 > res.length()) return;
        res = min(res, process());
    }
    String min(String a, String b)
    {
        if(a == null) return b;
        if(a.length() != b.length()) return a.length() < b.length() ? a : b;
        return a.compareTo(b) <= 0 ? a : b;
    }
    String process()
    {
        StringBuilder sb = new StringBuilder();
        Queue<Integer> qx = new LinkedList<Integer>();
        Queue<Integer> qy = new LinkedList<Integer>();
        qx.offer(sx);
        qy.offer(sy);
        int dist = d[sx][sy];
        while(dist-- > 1)
        {
            char best = 'z';
            Queue<Integer> qa = new LinkedList<Integer>();
            Queue<Integer> qb = new LinkedList<Integer>();
            boolean[][] vs = new boolean[rows][cols];
            while(!qx.isEmpty())
            {
                int cx = qx.poll(), cy = qy.poll();
                
                for(int k = 0; k < 4; k++)
                {
                    int tx = cx + dx[k];
                    int ty = cy + dy[k];
                    
                    if(tx < 0 || tx >= rows || ty < 0 || ty >= cols) continue;
                    if(vs[tx][ty]) continue;
                    vs[tx][ty] = true;
                    if(d[tx][ty] != dist) continue;
                    
                    qa.offer(tx);
                    qb.offer(ty);
                    if(cs[tx][ty] < best) best = cs[tx][ty];
                }
            }
            sb.append(best);

            while(!qa.isEmpty())
            {
                int x = qa.poll(), y = qb.poll();
                
                if(cs[x][y] == best)
                {
                    qx.offer(x);
                    qy.offer(y);
                }
            }
        }
        return sb.toString();
    }
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
    int INF = Integer.MAX_VALUE / 3;
    void BFS(int x, int y, int mask)
    {
        Queue<Integer> qx = new LinkedList<Integer>();
        Queue<Integer> qy = new LinkedList<Integer>();
        qx.offer(x);
        qy.offer(y);
        d = new int[rows][cols];
        for(int i = 0; i < rows; i++) for(int j = 0; j < cols; j++) d[i][j] = INF;
        d[x][y] = 0;
        while(!qx.isEmpty())
        {
            int cx = qx.poll(), cy = qy.poll();
            
            for(int k = 0; k < 4; k++)
            {
                int tx = cx + dx[k];
                int ty = cy + dy[k];
                
                if(tx < 0 || tx >= rows || ty < 0 || ty >= cols) continue;
                if(d[tx][ty] != INF) continue;
                int val = cs[tx][ty] - 'a';
                
                if(cs[tx][ty] == 'S' || (mask >> val & 1) != 0)
                {
                    d[tx][ty] = d[cx][cy] + 1;
                    qx.offer(tx);
                    qy.offer(ty);
                }
            }
        }
    }
    
    String next() throws IOException
    {
        while(st == null || !st.hasMoreTokens()) st = new StringTokenizer(input.readLine());
        return st.nextToken();
    }
    
    int nextInt() throws IOException
    {
        return Integer.parseInt(next());
    }
    
    long nextLong() throws IOException
    {
        return Long.parseLong(next());
    }
    
    double nextDouble() throws IOException
    {
        return Double.parseDouble(next());
    }
    
    void print(Object... o)
    {
        System.out.println(deepToString(o));
    }
    
    void gcj(Object o)
    {
        String s = String.valueOf(o);
        out.println("Case #" + test + ": " + s);
        System.out.println("Case #" + test + ": " + s);
    }
    
    BufferedReader input;
    PrintWriter out;
    StringTokenizer st;
    int test;
    
    void init() throws IOException
    {
        if(TEST) input = new BufferedReader(new FileReader(file + ".in")); 
        else input = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(new BufferedOutputStream(System.out));
    }
    
    public static void main(String[] args) throws IOException
    {
        new Thread(null, new C(), "", 1 << 20).start();
    }
    
    public void run()
    {
        try
        {
            init();
            if(TEST) 
            {
                int runs = nextInt();
                for(int i = 0; i < runs; i++) solve();
            }
            else solve();
            out.close();        
        }
        catch(Exception e)
        {
            e.printStackTrace();
            System.exit(1);
        }
    }
}