//https://cses.fi/problemset/task/1707
#include <bits/stdc++.h>
using namespace std;
/*
有点像793的题目，两次最短。Bee走出到每一个位置的最短，
然后二分答案，然后Mecho走最短的时侯每一个相邻是否可以走需要计算是否Bee先到。
如何处理一分钟走s步这里？dis记录当前的步数，每次走到相邻+1，但是时间可以用dis/s得到。
注意：一个位置如果bee先到，比如2s到，m 3s到，那么m就不能走，如果相等可以走，因为每一次都是m先走。
但是，终点要特殊处理。 能到达终点的周围四个点就可以到达。
*/
int n,s;
char a[801][801];
int rdir[4] = {0,0,1,-1};
int cdir[4] = {1,-1,0,0};
vector<vector<int>> beetime(801,vector<int>(801,1e9));
int sx,sy,ex,ey;//起点和终点
bool isok(int eat)//eat is how many time he stay at start point.
{
if (eat >= beetime[sx][sy]) return false;
vector<vector<int>> mstep(801,vector<int>(801,1e9));
//cout << eat << endl;
queue<pair<int,int>> myq;
mstep[sx][sy] = 0;
myq.push({sx,sy});
while (!myq.empty()) {
int x = myq.front().first;
int y = myq.front().second;
myq.pop();
for (int i = 0; i < 4; i++)
{
int newx = x + rdir[i];
int newy = y + cdir[i];
if (a[newx][newy] == 'D') return true;//终点不需要判断了。因为bee到不了。
if (newx <0 || newx == n || newy < 0 || newy == n) continue;
if (a[newx][newy] == 'T' || a[newx][newy] == 'H' ) continue;
if (mstep[newx][newy] <= mstep[x][y] + 1) continue;
//if (beetime[newx][newy] < ceil(1.0*(mstep[x][y]+1) /s) + eat) continue;
if (beetime[newx][newy] <= (mstep[x][y]+1) /s + eat) continue;
//cout << newx << " " << newy << " " << (mstep[x][y]+1) /s + eat << endl;
mstep[newx][newy] = mstep[x][y] + 1;
myq.push({newx,newy});
}
}
//cout << mstep[2][2] << endl;
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//freopen("nocross.in","r",stdin);
//freopen("nocross.out","w",stdout);
cin >> n >> s;
queue<pair<int,int>> myq;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
{
cin >> a[i][j];
if (a[i][j] == 'M') sx = i, sy = j;
if (a[i][j] == 'D') ex = i, ey = j;
if (a[i][j] == 'H') {
beetime[i][j] = 0;
myq.push({i,j});
};
}
//find beetime
while (!myq.empty()) {
int x = myq.front().first;
int y = myq.front().second;
myq.pop();
for (int i = 0; i < 4; i++)
{
int newx = x + rdir[i];
int newy = y + cdir[i];
if (newx <0 || newx == n || newy < 0 || newy == n) continue;
if (a[newx][newy] == 'T' || a[newx][newy] == 'H' || a[newx][newy] == 'D') continue;
if (beetime[newx][newy] <= beetime[x][y] + 1) continue;
beetime[newx][newy] = beetime[x][y] + 1;
myq.push({newx,newy});
}
}
// cout << beetime[2][2]<< endl;
//bin the answer
int ans = -1;
int l = 0, r = n*n, mid;
while (l <= r) {
mid = (l+r)/2;
//cout << "mid " << mid << endl;
if (isok(mid)) {
ans = mid;
l = mid + 1;
}
else r = mid - 1;
}
cout << ans << '\n';
return 0;
}
