首先按宽度w的奇偶来分情况考虑，解决方向肯定不是消去全部方块然后再还原，而应该是构造一个方案使加进来的俄罗斯方块不会对原图有影响。

然后就是伤脑的画图：

图很快就画好，但实现有好多case :)，我的实现方式要特判$4$的情况，因为w-4等于0！！！

代码：

#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

int n,w,a[30][2009];
char c;

void solve1(){
	int i=1;
	for(;!a[(i-1)/w+1][(i-1)%w+1]&&i<=w*n;++i);
	i=(i>w*n)?1:((i-1)%w+1);
	int l=min(max(0,i-2),w-3),r=min(max(0,n-i-1),w-3);
	if(l&1){
		cout<<4+l/2+(w-1-(l+2))/2+(w-2)/2<<"\n";
		cout<<"T 2 "<<l+1<<"\n";
		for(int j=l-1;j>0;j-=2){
			cout<<"S 0 "<<j<<"\n";
		}
		for(int j=l+3;j+2<=w;j+=2){
			cout<<"Z 0 "<<j<<"\n";
		}
		{
			cout<<"T 3 1\nT 1 "<<w-1<<"\n";
		}
		for(int j=1;j+4<=w;j+=2)cout<<"Z 0 "<<j<<"\n";
		cout<<"T 0 "<<w-2<<"\n";
	}else{
		cout<<4+l/2+(w-1-(l+2))/2+(w-5)/2<<"\n";
		cout<<"T 2 "<<l+1<<"\n";
		for(int j=l-1;j>0;j-=2){
			cout<<"S 0 "<<j<<"\n";
		}
		for(int j=l+3;j+2<=w;j+=2){
			cout<<"Z 0 "<<j<<"\n";
		}
		{
			cout<<"T 3 1\nT 1 "<<w-1<<"\n";
		}
		for(int j=2;j+5<=w;j+=2)cout<<"Z 0 "<<j<<"\n";
		cout<<"T 0 "<<w-3<<"\n";
	}

}

void solve2(){
	int i=1;
	for(;!a[(i-1)/w+1][(i-1)%w+1]&&i<=w*n;++i);
	i=i>w*n?1:((i-1)%w+1);
	int l=min(max(0,i-2),w-3),r=min(max(0,n-i-1),w-3);
	if(l&1){
		cout<<1+l/2+((w-1)-(l+2))/2+3+(w-1-4)/2+1<<"\n";
	}else cout<<1+l/2+((w-1)-(l+2))/2+3+(w-1-4)/2+1<<"\n";
	cout<<"T 2 "<<l+1<<"\n";
	for(int j=l-1;j>0;j-=2){
		cout<<"S 0 "<<j<<"\n";
	}
	for(int j=l+3;j+2<=w;j+=2){
		cout<<"Z 0 "<<j<<"\n";
	}
	if(l&1){
		cout<<"Z 1 1\n";
		cout<<"T 2 3\n";
		cout<<"L 1 1\n";
		for(int j=5;j+2<=w;j+=2){
			cout<<"Z 0 "<<j<<"\n";
		}
		cout<<"L 2 "<<w-1<<"\n";
	}else{
		cout<<"S 1 "<<w-1<<"\n";
		cout<<"T 2 "<<w-4<<"\n";
		cout<<"J 3 "<<w-2<<"\n";
		for(int j=w-6;j>0;j-=2)cout<<"S 0 "<<j<<"\n";
		cout<<"J 2 1\n";
	}
}

int main()
{
    scanf("%d%d",&w,&n);
    for(int i=1;i<=n;++i){
    	for(int j=1;j<=w;++j){
    		scanf(" %c",&c);
    		a[i][j]=c=='#';
    	}
    }
    if(w==4){
    	cout<<"1\nI 1 1\n";
    	return 0;
    }
    if(w&1)solve1();
    else solve2();
    return 0;
}