Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.

The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].

You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.

Input

The input consists of only one line. The only line contains the string s (1 ≤ |s| ≤ 106). Each character in s will be either "l" or "r".

Output

Output n lines — on the i-th line you should print the i-th stone's number from the left.

Examples

Input

llrlr


Output

3
5
4
2
1


Input

rrlll


Output

1
2
5
4
3


Input

lrlrr


Output

2
4
5
3
1

Note

In the first example, the positions of stones 1, 2, 3, 4, 5 will be <image>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
import java.util.*;
import static java.lang.System.*;

public class C264 {
    Scanner sc = new Scanner(in);

    public void run() {
        LinkedList<Integer> s=new LinkedList<Integer>();

        char[] lr=sc.nextLine().toCharArray();
        for(int i=lr.length-1;i>=0;i--){
            if(lr[i]=='l'){
                s.addLast(i+1);
            }else{
                s.addFirst(i+1);
            }
        }
        StringBuilder sb=new StringBuilder();
        for(int i :s){
            sb.append(i+"\n");
        }
        pr(sb);
    }

    public static void main(String[] _) {
        new C264().run();
    }

    public static void pr(Object o) {
        out.print(o);
    }
    public static void ln(Object o) {
        out.println(o);
    }
    public static void ln() {
        out.println();
    }
}