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.util.*;
import java.lang.*;
import java.io.*;   
import java.math.*;
public class Prac{     
    static class InputReader { 
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar, snumChars;
        public InputReader(InputStream st) {
            this.stream = st;
        } 
        public int read() {
            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } 
                catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }
        public int ni() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public long nl() {
            int c = read();
            while (isSpaceChar(c)) {
            c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public int[] nia(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public int[] nia1(int n) {
            int a[] = new int[n+1];
            for (int i = 1; i <=n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public long[] nla(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        } 
        public long[] nla1(int n) {
            long a[] = new long[n+1];
            for (int i = 1; i <= n; i++) {
                a[i] = nl();
            }
            return a;
        } 
        public Long[] nLa(int n) {
            Long a[] = new Long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        } 
        public Long[] nLa1(int n) {
            Long a[] = new Long[n+1];
            for (int i = 1; i <= n; i++) {
                a[i] = nl();
            }
            return a;
        } 
        public Integer[] nIa(int n) {
            Integer a[] = new Integer[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public Integer[] nIa1(int n) {
            Integer a[] = new Integer[n+1];
            for (int i = 1; i <= n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public String rs() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                    c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        } 
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        } 
    }
    public static class Key {
        private final int x, y;
        public Key(int x, int y) {
            this.x = x; this.y = y;
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof Key)) return false;
            Key key = (Key) o;
            return x == key.x && y == key.y;
        }   
        @Override
        public int hashCode() {
            int result = x;
            result = 31 * result + y;
            return result;
        }
    }
    static class Pair{
            int x,y;
            Pair(int a,int b){
                    x=a;y=b;
            }
//            @Override
//            public int compareTo(Pair p) {
//                    if(x==p.x) return y-p.y;
//                    return x-p.x;
//            }
    }
    static void shuffleArray(int temp[]){
        int n = temp.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            int tmp = temp[i];
            int randomPos = i + rnd.nextInt(n-i);
            temp[i] = temp[randomPos];
            temp[randomPos] = tmp;
        }   
    }
    static long gcd(long a,long b){ return b==0?a:gcd(b,a%b);}
    static long lcm(long a,long b){return (a/gcd(a,b))*b;}
    static PrintWriter ww = new PrintWriter(System.out);
    static long mod=998244353L,mod1=1000000007;
    //static int r[]={0,1,0,-1}, c[]={1,0,-1,0};
    static long modInverse(long a, long m){
        long num=m;
        long x=1;
        long power=a%mod;
        while(num>0){
              if(num%2==1){
                  x=(x*power)%mod;
            }
            num>>=1;
            power=(power*power)%mod;         
        }
        return x;
    }
    static long fact[]=new long[1000005];
    public static void findFact(){
        fact[0]=1;
        for(int i=1;i<=1000000;i++){
            fact[i]=(i*fact[i-1])%mod;
        }
    } 
    static long ncrWithMod(int n,int r){
        if(n<r)return 0L;

        return (fact[n]*modInverse((fact[r]*fact[n-r])%mod,mod-2))%mod;
    }

    public static void main(String [] args){
        InputReader sc=new InputReader(System.in); 
        String s = sc.nextLine();
        int n = s.length();
        ww.println(3);
        ww.println("R " + (n - 1));
        n++;
        ww.println("L " + (n - 1) );
        n += (n - 2);
        ww.println("L "+2);
        
        ww.close();
    }
}