#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int MAXN = 2001000;
char s[MAXN];
list<int> l;
int n;
int main() {
  scanf(" %s", s);
  n = (int)strlen(s);
  list<int>::iterator ptr = l.begin();
  for (int i = 0; i < (int)(n); i++) {
    ptr = l.insert(ptr, i);
    if (s[i] == 'r') ++ptr;
  }
  for (list<int>::iterator i = l.begin(); i != l.end(); i++)
    printf("%d ", (*i) + 1);
  cerr << clock() << endl;
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int MAXN = 2001000;
char s[MAXN];
list<int> l;
int n;
int main() {
  scanf(" %s", s);
  n = (int)strlen(s);
  list<int>::iterator ptr = l.begin();
  for (int i = 0; i < (int)(n); i++) {
    ptr = l.insert(ptr, i);
    if (s[i] == 'r') ++ptr;
  }
  for (list<int>::iterator i = l.begin(); i != l.end(); i++)
    printf("%d ", (*i) + 1);
  cerr << clock() << endl;
  return 0;
}

```