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
/**
 * Problem: 
 * Source: ', Round
 * Link: 
 * 
 * @author Alexei Ostrovski
 */

import java.io.*;
import java.util.*;


public class HighwaySolver {

	public static final int N_CHARS = 26;
	
	public static int n, m;
	public static char[][] field;
	public static int x_s, y_s, x_t, y_t;
	
	public static final int MAX_VALUE = 20000; //bigger than any possible result
	public static int resLength = MAX_VALUE;
	public static String resString = "";
	
	public static final boolean DEBUG = false;
	
	public static void main(String[] args) {
		//redirect input and output
		if (DEBUG)
		{
			try {
				System.setIn(new FileInputStream("input.txt"));
				//System.setOut(new PrintStream("input.txt"));
			} catch (IOException e) {
				//nothing
			}
			//generate();
			//return;
		}
		Scanner sc = new Scanner(System.in);

		//read data
		n = sc.nextInt(); m = sc.nextInt(); int k = sc.nextInt();
		field = new char[n][m];
		
		for (int i = 0; i < n; i++)
		{
			String line = sc.next();
			field[i] = line.toCharArray();
		}
		
		//solve
		//find all unique chars in the field
		String chars = "";
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
			{
				if (field[i][j] == 'S')
				{
					x_s = (int)i;	 y_s = (int)j;
				}
				if (field[i][j] == 'T')
				{
					x_t = (int)i;	 y_t = (int)j;
				}
				
				if ((field[i][j] != 'S') && (field[i][j] != 'T') &&
						(chars.indexOf(field[i][j]) < 0))
				{
					chars += field[i][j];
				}
			}
		
		//check if start and terminal cells are adjacent
		for (int i = 0; i < 4; i++)
			if ((x_s + dx[i] == x_t) && (y_s + dy[i] == y_t))
			{
				System.out.println();
				return;
			}
		
		char[] uniqueChars = chars.toCharArray();
		Arrays.sort(uniqueChars);
		
		if (k > uniqueChars.length)
			k = uniqueChars.length;
		//kinda clumsy, but anyway...
		/*if (k == 0) { 
			System.out.println();
			return;
		}
		else*/ if (k == 1)
		{
			for (int i1 = 0; i1 < uniqueChars.length; i1++)
				path("" + uniqueChars[i1]);
		} else if (k == 2) {
			for (int i1 = 0; i1 < uniqueChars.length; i1++)
				for (int i2 = i1 + 1; i2 < uniqueChars.length; i2++)
					path("" + uniqueChars[i1] + uniqueChars[i2]);
		} else if (k == 3) {
			for (int i1 = 0; i1 < uniqueChars.length; i1++)
				for (int i2 = i1 + 1; i2 < uniqueChars.length; i2++)
					for (int i3 = i2 + 1; i3 < uniqueChars.length; i3++)
						path("" + uniqueChars[i1] + uniqueChars[i2] + uniqueChars[i3]);
		} else { //k == 4
			for (int i1 = 0; i1 < uniqueChars.length; i1++)
				for (int i2 = i1 + 1; i2 < uniqueChars.length; i2++)
					for (int i3 = i2 + 1; i3 < uniqueChars.length; i3++)
						for (int i4 = i3 + 1; i4 < uniqueChars.length; i4++)
							path("" + uniqueChars[i1] + uniqueChars[i2] + uniqueChars[i3] +
									uniqueChars[i4]);
		}
		
		//output
		if (resLength < MAX_VALUE)
			System.out.println(resString);
		else
			System.out.println(-1);
	}
	
	public static void generate()
	{
		int n = 50, m = 50, k = 4;
		System.out.println(n + " " + m + " " + k);
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
				if ((i == 0) && (j == 0))
					System.out.print("S");
				else if ((i == n - 1) && (j == m - 1))
					System.out.print("T");
				else if ((i == n / 2) && (j == m / 2))
					System.out.print("a");
				else
					System.out.print("b");
			System.out.println();
		}
	}
	
	public static final int[] dx = {-1, 0, 0, 1};
	public static final int[] dy = {0, -1, 1, 0};

	//public static int cc = 0;
	
	public static void path(String allowedChars)
	{

		//System.out.println(allowedChars);
		//check allowed chars
		boolean near = false;
		for (int i = 0; i < 4; i++)
			if ((x_s + dx[i] >= 0) && (x_s + dx[i] < n) &&
					(y_s + dy[i] >= 0) && (y_s + dy[i] < m) &&
					(allowedChars.indexOf(field[x_s + dx[i]][y_s + dy[i]]) >= 0))
				near = true;
		if (!near)	return;
		near = false;
		for (int i = 0; i < 4; i++)
			if ((x_t + dx[i] >= 0) && (x_t + dx[i] < n) &&
					(y_t + dy[i] >= 0) && (y_t + dy[i] < m) &&
					(allowedChars.indexOf(field[x_t + dx[i]][y_t + dy[i]]) >= 0))
				near = true;
		if (!near)	return;
		
		int[][] queue = new int[m*n][4];
		int[][] l = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				l[i][j] = -1;
		boolean[][][] adj = new boolean[n][m][4];
		
		int len = 1, cur = 0;
		queue[0][0] = x_t;	//coordinates
		queue[0][1] = y_t;
		queue[0][2] = 0;	//length of path
		queue[0][3] = -1;	//parent
		l[x_t][y_t] = 0;

		while (cur < len)
		{
			int x = queue[cur][0], y = queue[cur][1];
			int curL = queue[cur][2];
			
			//System.out.println("(" + x + ", " + y + ") " + curL);
			if (curL > resLength + 1) //result cannot be improved
				return;
			if ((x == x_s) && (y == y_s)) //starting point reached
				break;
				
			for (int i = 0; i < 4; i++)
				if ((x + dx[i] >= 0) && (x + dx[i] < n) &&
						(y + dy[i] >= 0) && (y + dy[i] < m) && 
						((allowedChars.indexOf(field[x + dx[i]][y + dy[i]]) >= 0) || 
						 (field[x + dx[i]][y + dy[i]] == 'S')))
				{
					if (l[x + dx[i]][y + dy[i]] < 0) //point isn't in queue
					{
						queue[len][0] = (int)(x + dx[i]);
						queue[len][1] = (int)(y + dy[i]);
						queue[len][2] = (int)(curL + 1);
						queue[len][3] = cur;
						l[x + dx[i]][y + dy[i]] = (int)(curL + 1);
						adj[x + dx[i]][y + dy[i]][3 - i] = true;
						len++;
					} else if (l[x + dx[i]][y + dy[i]] == curL + 1) {
						adj[x + dx[i]][y + dy[i]][3 - i] = true;
					}
				}
			cur++; 
		}

		boolean less = (queue[cur][2] - 1 < resLength);
		//System.out.println(Arrays.toString(adj[1][0]));
		if ((cur < len) && (queue[cur][0] == x_s) && (queue[cur][1] == y_s) && 
				(queue[cur][2] - 1 <= resLength)) 
		{ //destination reached
			
			Set<Integer> set = new TreeSet<Integer>();
			//Set<int> newSet = new TreeSet<int>();
			//elements are stored in form x*n+y
			set.add((int)(x_s*m + y_s));
			
			String pathStr = "";
			
			for (int t = 0; t < queue[cur][2] - 1; t++)
			{
				char minChar = 'z' + 1;
				int x, y;
				
				for (Iterator<Integer> iter = set.iterator(); iter.hasNext(); )  
				{
					int val = iter.next();
					x = (int)(val / m);	
					y = (int)(val % m);
					for (int i = 0; i < 4; i++)
						if ((adj[x][y][i]) && (field[x + dx[i]][y + dy[i]] < minChar))
							minChar = field[x + dx[i]][y + dy[i]];					
				}
				if ((queue[cur][2] - 1 == resLength) && (!less))
				{
					if (minChar < resString.charAt(t))
						less = true;
					else if (minChar > resString.charAt(t))
					{ //result cannot be improved
						return;
					}
				}
				pathStr += minChar;				
				
				Set<Integer> newSet = new TreeSet<Integer>();
				//newSet.clear();
				for (Iterator<Integer> iter = set.iterator(); iter.hasNext(); )  
				{
					int val = iter.next();
					x = (int)(val / m);	
					y = (int)(val % m);					
					for (int i = 0; i < 4; i++)
						if ((adj[x][y][i]) && (field[x + dx[i]][y + dy[i]] == minChar)) //!!!
						{
							newSet.add((int)((x + dx[i])*m + (y + dy[i])));						
						}
				}
				
				set = newSet;
			}
			//System.out.println(allowedChars + ": " + pathStr);
			
			/*if (queue[cur][2] - 1 < resLength) {
				resString = pathStr;
				resLength = queue[cur][2] - 1; 
			} else if (pathStr.compareTo(resString) < 0)
				resString = pathStr;*/
			if ((pathStr.length() < resLength) || ((pathStr.length() == resLength) && 
					(less)))
			{
				resString = pathStr;
				resLength = queue[cur][2] - 1;//resString.length();
			}
			//System.out.println(resLength);
		}
	}	
}
