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.math.BigInteger;
import java.util.*;


public class Equal {
    static class pair implements Comparable<pair> {
        int v, w;

        public pair(int v, int w) {
            this.v = v;
            this.w = w;
        }

        @Override
        public int compareTo(pair o) {
            if (v == o.v)
                return w - o.w;
            return o.v - v;
        }
    }

    static class tri implements Comparable<tri> {
        int v, i, j;

        public tri(int v, int i, int j) {
            this.v = v;
            this.i = i;
            this.j = j;
        }

        @Override
        public int compareTo(tri o) {

            return v - o.v;
        }
    }

    static ArrayList<Integer>[] tree;
    static int timer = 0;
    static int[] timein;
    static int[] timeout;
    static int[][] powerParent;
    static int[] depth;
    static int[][] minWeight;

    static void dfs(int u, int parent) {

        timein[u] = ++timer;
//        powerParent[u][0] = parent;
////        minWeight[u][0] = weight;
//        if (parent == -1)
//            powerParent[u][0] = u;
//        for (int i = 1; i <= 32; i++) {
//            powerParent[u][i] = powerParent[powerParent[u][i - 1]][i - 1];
//            minWeight[u][i] = Math.min(minWeight[u][i - 1], minWeight[powerParent[u][i - 1]][i - 1]);
//        }
        for (int i = 0; i < tree[u].size(); i++) {
            int v = tree[u].get(i);
            if (v == parent)
                continue;
            depth[v] = depth[u] + 1;
            dfs(v, u);
        }


        timeout[u] = timer;
    }

    static boolean isAncestor(int u, int v) {
        return timein[u] <= timein[v] && timeout[v] <= timeout[u];
    }

    static int lca(int u, int v) {
        if (isAncestor(u, v))
            return u;
        if (isAncestor(v, u))
            return v;
        for (int i = 32; i >= 0; i--) {
            int temp = powerParent[u][i];
            if (!isAncestor(temp, v)) {
                u = temp;
            }
        }
        return powerParent[u][0];
    }

    static int distance(int u, int v) {
        return depth[u] + depth[v] - 2 * depth[lca(u, v)];
    }

    static int nthAncestor(int u, int n) {
        int c = -1;

        while (n > 0) {
            c++;
            if (n % 2 == 0) {
                n /= 2;
                continue;
            }
            u = powerParent[u][c];
            n /= 2;

        }
        return u;
    }

    static int nthMinWeight(int u, int n) {
        int c = -1;

        int min = (int) (1e9);
        while (n > 0) {
            c++;
            if (n % 2 == 0) {
                n /= 2;
                continue;
            }
            min = Math.min(min, minWeight[u][c]);
            u = powerParent[u][c];
            n /= 2;

        }
        return min;
    }


    static class SegmentTree {        // 1-based DS, OOP

        int N;            //the number of elements in the array as a power of 2 (i.e. after padding)
        int[] array, lazy;
        double[] sTree;

        public SegmentTree(int[] in) {
            array = in;
            N = in.length - 1;
            sTree = new double[N << 1];        //no. of nodes = 2*N - 1, we add one to cross out index zero
            lazy = new int[N << 1];
            build(1, 1, N);
        }

        public void build(int node, int b, int e)    // O(n)
        {
            if (b == e)
                sTree[node] = array[b];
            else {
                int mid = b + e >> 1;
                build(node << 1, b, mid);
                build(node << 1 | 1, mid + 1, e);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
            }
        }


        public void update_point(int index, double oval, double nval)            // O(log n)
        {
            index += N - 1;
            sTree[index] -= oval;
            sTree[index] += nval;
            while (index > 1) {
                index >>= 1;
                sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
            }
        }


        public void update_range(int i, int j, double oval, double nval)        // O(log n)
        {
            update_range(1, 1, N, i, j, oval, nval);
        }

        public void update_range(int node, int b, int e, int i, int j, double oval, double nval) {
            if (i > e || j < b)
                return;
            if (b >= i && e <= j) {
                sTree[node] -= oval;
                sTree[node] += nval;

            } else {
                int mid = b + e >> 1;
//                propagate(node, b, mid, e);
                update_range(node << 1, b, mid, i, j, oval, nval);
                update_range(node << 1 | 1, mid + 1, e, i, j, oval, nval);
                sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
            }

        }

//        public void propagate(int node, int b, int mid, int e) {
//            lazy[node << 1] += lazy[node];
//            lazy[node << 1 | 1] += lazy[node];
//            sTree[node << 1] += (mid - b + 1) * lazy[node];
//            sTree[node << 1 | 1] += (e - mid) * lazy[node];
//            lazy[node] = 0;
//        }

        public double query(int i, int j) {
            return query(1, 1, N, i, j);
        }

        public double query(int node, int b, int e, int i, int j)    // O(log n)
        {
            if (i > e || j < b)
                return 0;
            if (b >= i && e <= j)
                return sTree[node];
            int mid = b + e >> 1;
//            propagate(node, b, mid, e);
            double q1 = query(node << 1, b, mid, i, j);
            double q2 = query(node << 1 | 1, mid + 1, e, i, j);
            return q1 + q2;

        }
    }

    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static int NumOfOccIn(char[] c) {
        int ans = 0;
        char[] pattern = {'a', 'b', 'a', 'c', 'a', 'b', 'a'};
        for (int i = 0; i < c.length; i++) {
            int j = i;
            for (; j < c.length; j++) {
                if (j == i + 7)
                    break;
                if (c[j] != pattern[j - i])
                    break;
            }
            if (j == i + 7)
                ans++;
        }
        return ans;
    }

    static char[] cc;

    static int NumOccAfterUpdate(char[] c, int start) {
        int ans = 0;
        cc = c.clone();
        char[] pattern = {'a', 'b', 'a', 'c', 'a', 'b', 'a'};
        for (int i = start; i < c.length; i++) {
            if (i == start + 7)
                break;
            if (cc[i] == '?') {
                cc[i] = pattern[i - start];
            }
            if (cc[i] == pattern[i - start])
                continue;
            else return 0;
        }
        return NumOfOccIn(cc);
    }

    static long pow(int a, int e) {
        long res = 1;
        while (e-- > 0)
            res *= a;
        return res;
    }

    public static void main(String[] args) throws IOException {
        //BufferedReader br = new BufferedReader(new FileReader("name.in"));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter out = new PrintWriter(System.out);
       String s=br.readLine();
       out.println(3);
       out.println("L 2");
       out.println("R 2");
       out.println("R "+((2*s.length())-1));

        out.flush();
        out.close();

    }


}