Problem: After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and m columns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:

  1. Blue towers. Each has population limit equal to 100. 
  2. Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side. 



Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).

Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.

He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks. 

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j-th character in the i-th line is '.' if you're allowed to build at the cell with coordinates (i, j) a tower (empty cell) or '#' if there is a big hole there. 

Output

Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.

Each of the following k lines must contain a single operation in the following format:

  1. «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y); 
  2. «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y); 
  3. «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y). 



If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.

Examples

Input

2 3
..#
.#.


Output

4
B 1 1
R 1 2
R 2 1
B 2 3


Input

1 3
...


Output

5
B 1 1
B 1 2
R 1 3
D 1 2
R 1 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 510;
const int mod = 998244353;
using namespace std;
int get_second(pair<int, int> i) { return i.second; }
long long power(long long x, long long y) {
  int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int trail_zero(int n) {
  while (n % 10 == 0) n /= 10;
  return n;
}
map<long long, long long> mc;
multiset<long long> second;
map<pair<long long, long long>, long long> m1;
vector<pair<int, int>> vxx;
vector<int> val(N);
map<char, int> m4;
map<char, int> m5;
char s111[N];
long long p2[N], p3[N];
long long n, k, m, x, q, y, ans1 = INT_MAX;
bool check(int a, int b) {
  if (a >= 0 || a < n) return true;
  if (b >= 0 || b < m)
    return true;
  else
    return false;
}
int idx[] = {0, -1, 0, 1};
int idy[] = {1, 0, -1, 0};
char a[505][505];
vector<vector<long long>> adj(N), ans[4];
vector<long long> vis(N), out(N), cnt(N);
bool visx[505][505];
vector<char> v1;
vector<pair<long long, long long>> v2;
map<long long, long long> mxx;
int xxx[111][111];
int used[100][100];
vector<pair<int, int>> e, o(100002);
pair<long long, long long> root;
bool prime_check(long long a) {
  if (a == 2) return true;
  for (long long i = 2; i <= sqrt(a); i++) {
    if (a % i) continue;
    return false;
  }
  return true;
}
int smallest_divisor(long long a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return i;
  }
  return a;
}
bool con(char c) {
  if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u')
    return true;
  else
    return false;
}
long long first[N], g[N], h[N];
long long az[3];
int checkr(int x, int y) {
  long long sum = 0;
  for (int i = x; i <= y; i++)
    for (int j = 1; j <= m; j++) sum += xxx[i][j];
  return sum;
}
int checkc(int x, int y) {
  long long sum = 0;
  for (int i = x; i <= y; i++)
    for (int j = 1; j <= n; j++) sum += xxx[j][i];
  return sum;
}
int max_ans = 0;
int t = 0, l = 0;
long long ind[100000], mx = 0;
vector<long long> color(N);
set<long long> s2[N];
inline bool in(int x, int y) {
  return 1 <= x && x < n + 1 && 1 <= y && y < m + 1;
}
void dfs(long long x, long long y) {
  visx[x][y] = 1;
  v1.emplace_back('B');
  v2.push_back({x, y});
  for (int i = 0; i < 4; i++) {
    long long c = x + idx[i];
    long long b = y + idy[i];
    if (in(c, b) && a[c][b] == '.' && !visx[c][b]) {
      dfs(c, b);
      v1.emplace_back('D');
      v2.push_back({c, b});
      v1.emplace_back('R');
      v2.push_back({c, b});
    }
  }
}
long long a3[N], v22[N];
string names[N];
int slidw[N];
map<string, int> m3, m2;
set<long long> s3;
vector<long long> v;
void brutedfs(long long x, long long y, long long z) {
  if (x > 10000000000) return;
  if (y == z) s3.insert(x);
  brutedfs(x * 10 + 4, y + 1, z);
  brutedfs(x * 10 + 7, y, z + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      if (a[i][j] == '.' && !visx[i][j]) dfs(i, j);
    }
  cout << v1.size() << "\n";
  for (int i = 0; i < v1.size(); i++)
    cout << v1[i] << " " << v2[i].first << " " << v2[i].second << "\n";
  return 0;
}