/**
 * @see 广搜 https://www.acwing.com/problem/content/174/
 */
#include "cstring"
#include "algorithm"
#include "queue"
#include "iostream"
#include "cstdlib"

using namespace std;
struct State {
 //line为长方体的状态,0:立着;1:横着躺,2:竖着躺
 int x, y, line;
};

const int N = 510;
int n, pass;
char g[N][N];

int dist[N][N][3];

bool check(int x, int y) {
 if (x < 0 || x >= n || y < 0 || y >= pass) {
 return true;
 }
 return g[x][y] == '#';

}

//从起始的3种状态,4个方向,操作后的状态
int d[3][4][3] = {
 {{-2, 0, 2}, {0, 1, 1}, {1, 0, 2}, {0, -2, 1}},
 {{-1, 0, 1}, {0, 2, 0}, {1, 0, 1}, {0, -1, 0}},
 {{-1, 0, 0}, {0, 1, 2}, {2, 0, 0}, {0, -1, 2}}
};

int bfs(State start, State end) {
 queue<State> queue;
 memset(dist, -1, sizeof dist);
 dist[start.x][start.y][start.line] = 0;
 queue.push(start);
 while (!queue.empty()) {
 auto t = queue.front();
 queue.pop();
 for (int i = 0; i < 4; i++) {
 State next = {t.x + d[t.line][i][0], t.y + d[t.line][i][1], d[t.line][i][2]};
 int x = next.x;
 int y = next.y;
 int line = next.line;
 if (check(x, y)) continue;
 if (line == 0) {
 if (g[x][y] == 'E') {
 continue;
 }
 } else if (line == 1) {
 if (check(x, y + 1)) {
 continue;
 }
 } else {
 if (check(x + 1, y)) {
 continue;
 }
 }
 if (dist[x][y][line] == -1) {
 dist[x][y][line] = dist[t.x][t.y][t.line] + 1;
 queue.push(next);
 }
 }
 }
 return dist[end.x][end.y][end.line];

}


int main() {
 while (cin >> n >> pass, n || pass) {
 for (int i = 0; i < n; i++) {
 scanf("%q", g[i]);
 }
 State start = {-1, -1, -1}, end;
 for (int i = 0; i < n; i++) {
 for (int j = 0; j < pass; j++) {
 if (g[i][j] == 'X' && start.x == -1) {
 if (g[i + 1][j] == 'X') {
 start = {i, j, 2};
 } else if (g[i][j + 1] == 'X') {
 start = {i, j, 1};
 } else {
 start = {i, j, 0};
 }
 } else {
 if (g[i][j] == 'O') {
 end = {i, j, 0};
 }
 }
 }
 }
 int res = bfs(start, end);

// cout << "起点" << start.x << " " << start.y << " " <<
// start.line << endl;
// cout << "终点" << end.x << " " << end.y << " " << end.line << endl;

 if (res == -1) {
 cout << "Impossible" << endl;
 } else {
 cout << res << endl;
 }
 }


}