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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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char a[N][N];
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int MIN = 0x3f3f3f3f;
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int n, m;
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/**
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* 功能:计算在白x行,蓝y行,红z行的情况下,修改的数量
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* @param x 白色行数
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* @param y 蓝色行数
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* @param z 红色行数
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* @return 修改的数量
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*/
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int check(int x, int y, int z) {
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//检查现在前x行白的个数
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int w = 0, r = 0, b = 0;
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int s = 0;
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//白的现有多少个?
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for (int i = 1; i <= x; i++)
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for (int j = 1; j <= m; j++)
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if (a[i][j] == 'W') w++;
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//修改量+ m*x-w个
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s += m * x - w;
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//蓝的现有多少个?
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for (int i = x + 1; i <= x + y; i++)
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for (int j = 1; j <= m; j++)
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if (a[i][j] == 'B') b++;
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//修改量+ m*y-b个
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s += m * y - b;
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//红的现有多少个?
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for (int i = x + y + 1; i <= x + y + z; i++)
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for (int j = 1; j <= m; j++)
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if (a[i][j] == 'R') r++;
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//修改量+ m*z-r个
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s += m * z - r;
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//返回当前情况下的修改量
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return s;
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}
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int main() {
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//n行m列的国旗
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cin >> n >> m;
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//读入现有的国旗
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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cin >> a[i][j];
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//白色,可能的范围是1行至n-2行(行数)
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for (int i = 1; i <= n - 2; i++)
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//蓝色,可能的范围是1行至n-2行(行数)
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for (int j = 1; j <= n - 2; j++) {
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//红色的行数
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int k = n - i - j;
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//如果红色不存在,那么就是没用的答案,j也没有必要再增大了,使用 continue无意义,直接break
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if (k <= 0) break;
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//在白i行,蓝j行,红k行的情况下,计算最小的修改量,取最后的最小值
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MIN = min(MIN, check(i, j, k));
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}
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//输出最小值
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cout << MIN << endl;
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return 0;
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}
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