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.util.*;

public class Track{

	static int[] dx = {-1, 1, 0, 0};
	static int[] dy = {0, 0, -1, 1};

	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int m = in.nextInt();
		int n = in.nextInt();
		int p = in.nextInt();
		char[][] chs = new char[m][];
		int[] S = null;
		int[] T = null;
		for(int i = 0; i < m; i++){
			String cur = in.next();
			chs[i] = cur.toCharArray();
			for(int j = 0; j < n; j++){
				if(chs[i][j] == 'S'){
					S = new int[] {i,j};
				}else if(chs[i][j] == 'T'){
					T = new int[] {i,j};
				}
			}
		}
		int best = 1<<30;
		String path = null;
		for(int i = 0; i < 26; i++){
			for(int j = i; j < 26; j++){
				for(int k = j; k < 26; k++){
					for(int l = k; l < 26; l++){
						boolean[] can = new boolean[26];
						if(p > 0){
							can[i] = true;
						}
						if(p > 1){
							can[j] = true;
						}
						if(p > 2){
							can[k] = true;
						}
						if(p > 3){
							can[l] = true;
						}
						int[][] d = new int[m][n];
						String[][] pat = new String[m][n];
						LinkedList<int[]> queue = new LinkedList<int[]>();
						queue.addLast(S);
						pat[S[0]][S[1]] = "";
						while(!queue.isEmpty()){
							int[] u = queue.removeFirst();
							for(int a = 0; a < 4; a++){
								int x = u[0]+dx[a];
								int y = u[1]+dy[a];
								if(0 <= x && x < m && 0 <= y && y < n){
									if(pat[x][y] == null && (chs[x][y] == 'T' || can[chs[x][y]-'a'])) {
										d[x][y] = d[u[0]][u[1]]+1;
										pat[x][y] = pat[u[0]][u[1]].concat(Character.toString(chs[x][y]));
										queue.addLast(new int[] {x,y});
									}else if(d[x][y] == d[u[0]][u[1]]+1) {
										String newPath = pat[u[0]][u[1]].concat(Character.toString(chs[x][y]));
										if(newPath.compareTo(pat[x][y]) < 0 ) {
											pat[x][y] = newPath;
										}
									}
								}
							}
							if(chs[u[0]][u[1]] == 'T') {
								break;
							}
						}
						if(path == null) {
							path = pat[T[0]][T[1]];
							best = d[T[0]][T[1]];
						}else if(pat[T[0]][T[1]] != null && d[T[0]][T[1]] <= best) {
							if(d[T[0]][T[1]] < best) {
								path = pat[T[0]][T[1]];
								best = d[T[0]][T[1]];
							}else if( pat[T[0]][T[1]].compareTo(path) < 0){
								path = pat[T[0]][T[1]];
								best = d[T[0]][T[1]];
							}
						}
					}
				}
			}
		}
		System.out.println(path != null ? path.substring(0, path.length()-1) : -1);
	}

}
