//分析:题意是要恰好在t时间内到达,所以只能用dfs进行回溯搜索(bfs搜索的是最短路)。
//若 t-[abs(ex-sx)+abs(ey-sy)] 结果为非偶数（奇数），则无法在t步恰好到达；

package 搜索;

import java.util.Scanner;

public class hdu_1010 {
    static char[][] map;
    static boolean flag;
    static boolean[][] vis;
    static int[][] dir = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

    static int n, m, t;
    static int sx, sy, ex, ey;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (true) {
            n = sc.nextInt();
            m = sc.nextInt();
            t = sc.nextInt();
            if (n == 0 && m == 0 && t == 0)
                break;
            map = new char[n][m];
            vis = new boolean[n][m];
            for (int i = 0; i < n; i++) {
                String str = sc.next();
                map[i] = str.toCharArray();
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (map[i][j] == ‘S‘) {
                        sx = i;
                        sy = j;
                    }
                    if (map[i][j] == ‘D‘) {
                        ex = i;
                        ey = j;
                    }
                }
            }
            if (t < Math.abs(ex - sx) + Math.abs(ey - sy)) {
                System.out.println("NO");
                continue;
            }
            if ((t - Math.abs(ex - sx) + Math.abs(ey - sy)) % 2 != 0) {
                System.out.println("NO");
                continue;
            }
            flag = false;
            vis[sx][sy]=true; //第一个点没标记WA了半天
            dfs(sx, sy, 0);
            if (!flag)
                System.out.println("NO");
            else
                System.out.println("YES");
        }
    }

    private static void dfs(int x, int y, int time) {
        if (x == ex && y == ey && time == t) {
            //System.out.println(x+" "+y + " "+time);
            flag = true;
            return;
        }
        if (time >= t) //时间过了
            return;
        for (int i = 0; i < 4; i++) {
            int next_x = x + dir[i][0];
            int next_y = y + dir[i][1];
            if (next_x < 0 || next_x >= n || next_y < 0 || next_y >= m
                    || vis[next_x][next_y] == true
                    || map[next_x][next_y] == ‘X‘) {
                continue;
            }
            vis[next_x][next_y] = true;
            dfs(next_x, next_y, time + 1);
            vis[next_x][next_y] = false; // 回溯
            if (flag)
                return; // 找到了直接返回
        }
    }
}