#include<iostream>
#include<cmath>
#include<cstring>

using namespace std;

#define N 15

char maps[N][N];

int n, m, t, flag, dx, dy, vis[N][N];
int dir[4][2] = { {0,1}, {0, -1}, {1, 0}, {-1, 0}};

void dfs(int x, int y, int k)
{
    if(flag) return ;

    if(x == dx && y == dy && k == 0) flag = 1;  // 如果正好到达，flag = 1；

    int tmp = abs(x - dx) + abs(y-dy); 

    if(k < tmp) return ;   // 如果所剩步数比所需步数 tmp小，return
 
    for(int i = 0; i < 4; i++)
    {
        int nx, ny;
        nx = x + dir[i][0]; 
        ny = y + dir[i][1];

        if(nx >= 0 && nx < n && ny >= 0 && ny < m && maps[nx][ny] != ‘X‘ && !vis[nx][ny])
        {
            vis[nx][ny] = 1; 
            dfs(nx, ny, k-1);
            vis[nx][ny] = 0;
        }
    }
}

int main()
{
    int sx, sy;
    while(cin >> n >> m >> t, n+m+t)
    {
        flag = 0; 
        memset(vis, 0, sizeof(vis));  // 每次都要清0；

        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
        {
            cin >> maps[i][j]; 

            if(maps[i][j] == ‘S‘) 
                sx = i, sy = j; 
            else if(maps[i][j] == ‘D‘)
                dx = i, dy = j; 
        }
        int q = (sx+sy)%2, w = (dx+dy)%2;    
        vis[sx][sy] = 1;
        if((q+w)%2 == t%2)     // 剪枝，如果到达所需步数和所给 t 奇偶性不一样的话，就不执行dfs了，no
            dfs(sx, sy, t); 
        if(flag)
            cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}