#include <bits/stdc++.h>
using namespace std;

int main() {
    int T;
    cin >> T;
    while (T--) {
        int n, m;
        cin >> n >> m;
        vector<string> grid(n);
        for (int i = 0; i < n; ++i) cin >> grid[i];

        bool all_white = true, all_black = true;
        int black = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 'W') all_black = false;
                else all_white = false;
                if (grid[i][j] == 'B') black++;
            }
        }

        // 如果已经全白或全黑
        if (all_white || all_black) {
            cout << "YES\n0\n";
            continue;
        }

        // 处理单行或单列的情况
        if (n == 1 || m == 1) {
            int total = n * m;
            if (total % 2 == 0) {
                // 总单元格数为偶数，只有初始全黑或全白才能通过0次操作，或者通过一次操作得到全白或全黑
                if (black == 0 || black == total) {
                    cout << "YES\n0\n";
                } else if (total - black == 0 || total - black == total) {
                    // 一次操作可以翻转所有单元格，得到全白或全黑
                    cout << "YES\n1\n";
                    if (n == 1) {
                        cout << string(m - 1, 'R') << '\n';
                    } else {
                        cout << string(n - 1, 'D') << '\n';
                    }
                } else {
                    cout << "NO\n";
                }
            } else {
                // 总单元格数为奇数，只有初始全白或全黑，或者一次操作后全白或全黑
                if (black == 0 || black == total) {
                    cout << "YES\n0\n";
                } else {
                    // 检查一次操作是否可以转为全白或全黑
                    if (total - black == 0) {
                        cout << "YES\n1\n";
                        if (n == 1) cout << string(m - 1, 'R') << '\n';
                        else cout << string(n - 1, 'D') << '\n';
                    } else if (total - black == total) {
                        cout << "YES\n1\n";
                        if (n == 1) cout << string(m - 1, 'R') << '\n';
                        else cout << string(n - 1, 'D') << '\n';
                    } else {
                        cout << "NO\n";
                    }
                }
            }
            continue;
        }

        // 处理行和列都大于1的情况，构造两条路径
        string path1 = string(m - 1, 'R') + string(n - 1, 'D');
        string path2 = string(n - 1, 'D') + string(m - 1, 'R');
        cout << "YES\n2\n";
        cout << path1 << '\n' << path2 << '\n';
    }
    return 0;
}