#include <bits/stdc++.h>

#define INF 1000000005

#define ll long long

using namespace std;



ifstream in("barbar.in");

ofstream out("barbar.out");



ll n,m,cost[1005][1005],st_y,st_x,end_x,end_y,viz[1005][1005];

ll dy[4] = {0 , 0 , -1 , 1};

ll dx[4] = {1 , -1 , 0 , 0};

char a[1005][1005];

queue <pair<ll,ll>> q;



void Fill(ll i,ll j,ll lim)

{

    viz[i][j] = 1;

    for (ll k = 0; k < 4; k++)

    {

        ll y = i + dy[k];

        ll x = j + dx[k];

        if (y >= 1 && y <= n && x >= 1 && x <= m)

        if ( (a[y][x] != '*') && cost[y][x] >= lim && viz[y][x] == 0)

            Fill(y , x , lim);

    }

}

bool ver(ll lim)

{

    ll i,j;



    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) viz[i][j] = 0;
	if (cost[st_y][st_x] >= lim)
    Fill(st_y , st_x , lim);
    else return false;





    if (viz[end_y][end_x] == 1) return true;

    return false;

}

ll solve(ll st, ll dr)

{

    ll mij,last = -1;

    bool f;

    while (st <= dr)

    {

        mij = (st + dr) / 2;

        f = ver(mij);

        if (f)

            last = mij,

            st = mij + 1;

        else

            dr = mij - 1;

    }

    return last;

}

int main()

{ll i,j;

    in >> n >> m;

    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) in >> a[i][j];

    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) cost[i][j] = INF;

    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) {

        if (a[i][j] == 'D') q.push({i,j}) , cost[i][j] = 0;

        else if (a[i][j] == 'I') st_y = i , st_x = j;

        else if (a[i][j] == 'O') end_y = i , end_x = j;

    }



    while (!q.empty())

    {

        i = q.front().first;

        j = q.front().second;

        q.pop();



        for (int k = 0; k < 4; k++)

        {

            ll y = i + dy[k];

            ll x = j + dx[k];



            if (cost[y][x] > cost[i][j] + 1 && a[y][x] != '*'){

                q.push({y,x});

                cost[y][x] = cost[i][j] + 1;

            }

        }

    }

    if (ver(0) == 0) out << "-1\n";

    else

    out << solve(0 , n * m) << "\n";

    out.close();

    in.close();

    return 0;

}