Problem: Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string. 

The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.

The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.

Note that characters in the string in this problem are indexed from 1.

For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.

Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6

It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.

Input

The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.

Output

The first line should contain k (0≤ k ≤ 30) — the number of operations performed.

Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.

The length of the resulting palindrome must not exceed 10^6.

Examples

Input


abac


Output


2
R 2
R 5


Input


acccc


Output


2
L 4
L 2


Input


hannah


Output


0

Note

For the first example the following operations are performed:

abac → abacab → abacaba

The second sample performs the following operations: acccc → cccacccc → ccccacccc

The third example is already a palindrome so no operations are required.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.nio.channels.ShutdownChannelGroupException;
public class Codeforces{
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long MOD = (long)(1e9+7);
	static long MOD2 = MOD*MOD;
	//static long MOD = 998244353;
	static FastReader sc = new FastReader();
	static int pInf = Integer.MAX_VALUE;
	static int nInf = Integer.MIN_VALUE;
	public static void main(String[] args){
		int test = 1;
		//test = sc.nextInt();
		while(test-->0)solve();
		out.flush();
		out.close();
	}
	public static void solve() {
		String s = sc.next();
		int n = s.length();
		out.println(3);
		out.println('R'+" "+(n-1));
		out.println('L'+" "+(n));
		out.println('L'+" "+2);
		
	}
	/*static class Node implements Comparable<Node>{
		int depth,subTreeSize,id;
		ArrayList<Node> adj = new ArrayList<>();
		public Node(int id) {
			this.id = id;
		}
		public void dfs0(Node par,int depth) {
			if(par!=null) {
				adj.remove(par);
			}
			this.depth = depth;
			this.subTreeSize = 1;
			for(Node nn:adj) {
				nn.dfs0(this, depth+1);
				subTreeSize += nn.subTreeSize;
			}
		}
		public int compareTo(Node o) {
			int value=depth-subTreeSize, oValue=o.depth-o.subTreeSize;
			return Integer.compare(value, oValue);
		}
		public String toString() {
			return "id: "+id+" depth: "+depth+" subtreeSize: "+subTreeSize; 
		}
	}*/
	public static int lowerBound(int key, ArrayList<Integer> A) {
		int s = 0;
		int e = A.size()-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(A.get(m)>=key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static int upperBound(int key, ArrayList<Integer> A) {
		int s = 0;
		int e = A.size()-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(A.get(m)>key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static long c2(long n) {
		if((n&1)==0) {
			return mul(n/2, n-1);
		}
		else {
			return mul(n, (n-1)/2);
		}
	}
	public static long mul(long a, long b){
	    return ((a%MOD)*(b%MOD))%MOD;
	}
	public static long add(long a, long b){
	    return ((a%MOD)+(b%MOD))%MOD;
	}
	public static long sub(long a, long b){
	    return ((a%MOD)-(b%MOD))%MOD;
	}
//Pair Class
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		int len;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
			this.len = y-x+1;
		}
		@Override
		public int compareTo(Pair o) {
			if(this.x==o.x){
			    return (this.y-o.y);
			}
			return -1*(this.x-o.x);
		}
	}
//nCr
	static long ncr(long n, long k) {
        long ret = 1;
        for (long x = n; x > n - k; x--) {
            ret *= x;
            ret /= (n - x + 1);
        }

        return ret;
    }
	static long finextDoubleMMI_fermat(long n,int M)
    {
        return fastExpo(n,M-2);
    }
 
    static long nCrModPFermat(int n, int r, int p) 
    { 
        if (r == 0) 
            return 1; 
        long[] fac = new long[n+1]; 
        fac[0] = 1;           
        for (int i = 1 ;i <= n; i++) 
            fac[i] = fac[i-1] * i % p;       
        return (fac[n]* finextDoubleMMI_fermat(fac[r], p)% p * finextDoubleMMI_fermat(fac[n-r], p) % p) % p; 
    }
//Shuffle Sort
    static final Random random=new Random();
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
//Merge Sort
	static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
//Brian Kernighans Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
//Factorial
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
//Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
//Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
//AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i<=Math.sqrt(n);i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
//Reverse an array
    static <T> void reverse(T arr[],int l,int r){
    	Collections.reverse(Arrays.asList(arr).subList(l, r));
    }
//Sieve of eratosthenes
    static int[] findPrimes(int n){
        boolean isPrime[]=new boolean[n+1];
        ArrayList<Integer> a=new ArrayList<>();
        int result[];
        Arrays.fill(isPrime,true);
        isPrime[0]=false;
        isPrime[1]=false;
        for(int i=2;i*i<=n;++i){
            if(isPrime[i]==true){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        for(int i=0;i<=n;i++) if(isPrime[i]==true) a.add(i);
        result=new int[a.size()];
        for(int i=0;i<a.size();i++) result[i]=a.get(i);
        return result;
        
    }
//Indivisual factors of all nos till n
    static ArrayList<Integer>[] indiFactors(int n){
    	ArrayList<Integer>[] A = new ArrayList[n+1];
    	for(int i = 0; i <= n; i++) {
    		A[i] = new ArrayList<>();
    	}
    	int[] sieve = new int[n+1];
    	for(int i=2;i<=n;i++) {
       		if(sieve[i]==0) {
       			for(int j=i;j<=n;j+=i) if(sieve[j]==0) {
       				//sieve[j]=i;
       				A[j].add(i);
       			}
       		}
       	}
    	return A;
    }
//Segmented Sieve
    static boolean[] segmentedSieve(long l, long r){
    	boolean[] segSieve = new boolean[(int)(r-l+1)];
    	Arrays.fill(segSieve, true);
    	int[] prePrimes = findPrimes((int)Math.sqrt(r));
    	for(int p:prePrimes) {
    		long low = (l/p)*p;
    		if(low < l) {
    			low += p;
    		}
    		if(low == p) {
    			low += p;
    		}
    		for(long j = low; j<= r; j += p) {
    			segSieve[(int) (j-l)] = false;
    		}
    	}
    	if(l==1) {
    		segSieve[0] = false;
    	}
    	return segSieve;
    }
//Euler Totent function
    static long countCoprimes(long n){
        ArrayList<Long> prime_factors=new ArrayList<>();
        long x=n,flag=0;
        while(x%2==0){
            if(flag==0) prime_factors.add(2L);
            flag=1;
            x/=2;
        }
        for(long i=3;i*i<=x;i+=2){
            flag=0;
            while(x%i==0){
                if(flag==0) prime_factors.add(i);
                flag=1;
                x/=i;
            }
        }
        if(x>2) prime_factors.add(x);
        double ans=(double)n;
        for(Long p:prime_factors){
            ans*=(1.0-(Double)1.0/p);
        }
        return (long)ans;
    }
	static long modulo = (long)1e9+7;
	public static long modinv(long x){
	    return modpow(x, modulo-2);
	}
	public static long modpow(long a, long b){
	    if(b==0){
	        return 1;
	    }
	    long x = modpow(a, b/2);
	    x = (x*x)%modulo;
	    if(b%2==1){
	        return (x*a)%modulo;
	    }
	    return x;
	}
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
}