#include <iostream>
#include <vector>
#include <deque>

using namespace std;

// 方向数组：右上，右下，左下，左上 (对应格点的移动)
const int dir[4][2] = {{-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
// 对应方向上，方格应该有的形状
const char expect[4] = {'/', '\\', '/', '\\'};
// 对应方向上，方格在 mat 中的坐标偏移
const int dr[4] = {-1, 0, 0, -1};
const int dc[4] = {0, 0, -1, -1};

const int INF = 1e9;

int main() {
    int n, m;
    if (!(cin >> n >> m)) return 0;

    // 奇偶性剪枝：如果终点坐标和为奇数，绝对无法到达
    if ((n + m) % 2 != 0) {
        cout << "NO SOLUTION" << endl;
        return 0;
    }

    vector<vector<char>> mat(n, vector<char>(m));
    // 修正：dist 必须是 int 类型，初始化为无穷大
    vector<vector<int>> dist(n + 1, vector<int>(m + 1, INF));

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

    deque<pair<int, int>> dq;
    dist[0][0] = 0;
    dq.push_back({0, 0});

    while (!dq.empty()) {
        pair<int, int> curr = dq.front();
        dq.pop_front();

        int r = curr.first;
        int c = curr.second;

        for (int i = 0; i < 4; ++i) {
            int nr = r + dir[i][0];
            int nc = c + dir[i][1];

            // 检查格点是否越界
            if (nr >= 0 && nr <= n && nc >= 0 && nc <= m) {
                // 对应方格的坐标
                int tr = r + dr[i];
                int tc = c + dc[i];
                
                // 如果当前方向的字符与地图不符，权值为 1，否则为 0
                int weight = (mat[tr][tc] == expect[i] ? 0 : 1);

                if (dist[r][c] + weight < dist[nr][nc]) {
                    dist[nr][nc] = dist[r][c] + weight;
                    if (weight == 0) {
                        dq.push_front({nr, nc}); // 权值为0，插到队头
                    } else {
                        dq.push_back({nr, nc});  // 权值为1，插到队尾
                    }
                }
            }
        }
    }

    if (dist[n][m] == INF) cout << "NO SOLUTION" << endl;
    else cout << dist[n][m] << endl;

    return 0;
}