#include <bits/stdc++.h>
#define fi first
#define II inline
#define se second
#define pii pair <int, int>
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)

using namespace std;

const int grid = 1002, INF = (int) 1e4 + 2;

char a[grid][grid], viz[grid][grid];

int dist[grid][grid];
pii paft; queue < pii > q;

II bool ispro (pii p, int N, int M){
    return (p.fi > 0 && p.se > 0 && p.fi <= N && p.se <= M);
}

II void dp (int N, int M){
    const int dx[] = {-1, 0, 1, 0};
    const int dy[] = {0, 1, 0, -1};
    for (; !q.empty(); q.pop()){
            pii v = q.front();
            for (int k = 0 ; k < 4 ; k++){
                    pii w = {v.fi + dx[k], v.se + dy[k]};
                       if (ispro(w, N, M) && dist[w.fi][w.se] > dist[v.fi][v.se] + 1 && a[w.fi][w.se] != '*'){
                            dist[w.fi][w.se] = dist[v.fi][v.se] + 1;
                            q.push(w);
                       }
            }
    }
}
II bool pos (int kthvalue, int N, int M){
    if (dist[paft.fi][paft.se] < kthvalue)
        return 0;
    const int dx[] = {-1, 0, 1, 0};
    const int dy[] = {0, 1, 0, -1};
    memset(viz, 0, sizeof(viz));
    q.push({paft.fi, paft.se});
    viz[paft.fi][paft.se] = 1;
    for (; !q.empty(); q.pop()){
            pii v = q.front();
            if (a[v.fi][v.se] == 'O') return 1;
            for (int k = 0 ; k < 4; k++){
                    pii w = {v.fi + dx[k], v.se + dy[k]};
                       if (ispro(w, N, M) &&  a[w.fi][w.se] != '*' && !viz[w.fi][w.se] && dist[w.fi][w.se] >= kthvalue){
                            viz[w.fi][w.se] = 1;
                       q.push(w);
                       }
            }
    }
    return 0;
}

II int solve (int N, int M){
    int left = 0, top = N * M, bstrez = INF;
    while (left <= top){
            int mid = left + ((top - left) >> 1 );
            if (pos(mid, N, M))
                left = mid + 1,bstrez = min(bstrez, mid);
            else top = mid - 1;
    }
    return (bstrez == INF ? -1 : bstrez);
}
int main(){
    ifstream cin ("barbar.in");
    ofstream cout ("barbar.out");
    memset(dist, INF, sizeof(dist) );
    int N, M;
    cin >> N >> M;
    FOR(i, 1, N)
       FOR(j, 1, M){
           cin >> a[i][j];
           if (a[i][j] == 'D') q.push({i, j}), dist[i][j] = 0;
              if (a[i][j] == 'I') paft = {i, j};
                 if (a[i][j] == '*') dist[i][j] = -1;

       }
       dp(N, M);
       cout << solve(N, M);

       return 0;
}