#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int state[200], num[200];
int cur[110];
char g[110][20];
int dp[110][200][200];
int n, m, top, total;
inline bool is_ok(int x)
{if (x & (x << 1))return 0;if (x & (x << 2))return 0;return 1;
}
int jcount(int x)
{int cnt = 0;for (int i = 0; i < 12; i++)if (x & (1 << i))cnt++;return cnt;
}
int fit(int x, int i)
{if (x & cur[i])return 0;return 1;
}
void init()
{top = 0;total = 1 << m;for (int i = 0; i < total; i++){if (is_ok(i))state[++top] = i;}
}
int main()
{while (scanf("%d%d", &n, &m) != EOF){init();for (int i = 1; i <= n; i++){cur[i] = 0;scanf("%s", g[i] + 1);for(int j = 1; j <= m; j++)if (g[i][j] == 'H')cur[i] += (1 << (j - 1) );
//同上一题一样将不能放炮的设为1，这样 & 的话只要非0就不行，因为一定含有不能放炮的而放炮了，0是允许放炮，但是放不放都行，也就是0,1随便，都是可以的
        }memset(dp, -1, sizeof(dp));int ans = -1;for (int i = 1; i <= top; i++){num[i] = jcount(state[i]);  // 求出每一种状态下能放得炮数if (!fit(state[i], 1))continue;dp[1][1][i] = num[i];  //第一行状态为state[i],第0行状态为state[1] = 0ans = max(ans, num[i]);}for (int i = 2; i <= n; i++){for (int t = 1; t <= top; t++){if (!fit(state[t], i))continue;for (int j = 1; j <= top; j++) // 第 i - 1行的情况
                {if (state[t] & state[j])continue;for (int k = 1; k <= top; k++) //第i - 2行的情况
                    {if (state[t] & state[k])continue;if (dp[i - 1][k][j] != -1)  // 不符合要求的，dp[1][1][其他] ,而第0行其他状态都没有
                        {dp[i][j][t] = max(dp[i][j][t], dp[i - 1][k][j] + num[t]);ans = max(ans, dp[i][j][t]);}}}}}printf("%d\n", ans);}return 0;
}