【CodeForces - 264A】Escape from Stones (模拟,卡精度的处理)

题干:

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, nstones 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 , respectively. So you should print the sequence: 3, 5, 4, 2, 1.

题目大意:

  有一只松鼠,他在【0,1】这个区间,然后有一堆石头从天而降,每次都是砸在他所在区间的中间,他会选择向左或者向右跳,问从左到右输出石头的编号。

解题报告:

  直接开long double 也会炸精度,,,这题其实同时放大几倍 就好了啊,因为一共有n个操作,所以就用n这么大的数组,然后模拟就行了。

 

AC代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 1e6 + 6;
char s[MAX];
int a[MAX];
int main() 
{cin>>(s+1);int len = strlen(s+1);int l = 1,r = len;for(int i = 1; i<=len; i++) {if(s[i] == 'l') a[l++]=i;else a[r--]=i;  } for(int i = len; i>=1; i--) printf("%d\n",a[i]);return 0;
}

 

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