题目来源：

https://vjudge.net/contest/293215#problem/C

http://codeforces.com/problemset/problem/939/E

You are given a multiset S consisting of positive integers (initially empty). There are two kind of queries:

  1. Add a positive integer to S, the newly added integer is not less than any number in it.
  2. Find a subset s of the set S such that the value is maximum possible. Here max(s) means maximum value of elements in s,  — the average value of numbers in s. Output this maximum possible value of .

Input

The first line contains a single integer Q (1 ≤ Q ≤ 5·105) — the number of queries.

Each of the next Q lines contains a description of query. For queries of type 1 two integers 1 and x are given, where x (1 ≤ x ≤ 109) is a number that you should add to S. It's guaranteed that x is not less than any number in S. For queries of type 2, a single integer 2 is given.

It's guaranteed that the first query has type 1, i. e. S is not empty when a query of type 2 comes.

Output

Output the answer for each query of the second type in the order these queries are given in input. Each number should be printed in separate line.

Your answer is considered correct, if each of your answers has absolute or relative error not greater than 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

Examples

Input

6
1 3
2
1 4
2
1 8
2

Output

0.0000000000
0.5000000000
3.0000000000

Input

4
1 1
1 4
1 5
2

Output

2.0000000000

题意：

第一行是一个Q，表示Q个操作。

操作有两种：

操作一：在集合S中添加元素x，保证x不小于S中任何一个元素（保证S递增）

操作二：查询，s为S子集，求函数max(s)-mean(s)的最大值，max(s)表示s中元素的最大值，mean(s)表示s中所有元素的平均数。对于每次操作二，输出对应函数最大值，误差小于10e-6

n为S元素个数，由于S递增，要使max(s)-mean(s)最大，只需要取S中的最后一个元素和S中前m(m<=n-1)个元素即可，而且由1到m函数max(s)-mean(s)的值先增加后减小

思路：三分最值+前缀和

三分原理：

lmid和rmid是l到r的两个三等分点，如果lmid大于rmid，那么最大值max一定在l与rmid之间，如果lmid小于rmid那么最大值max一定在lmid与r之间。

三分讲解具体参考：https://blog.csdn.net/forever_dreams/article/details/81093369

比赛的时候就想到是三分，但是不会实现，看见大佬们都没ac所以直接放弃了尝试，，，其实贼简单，没人做大概是题目没读懂吧（翻译好评Orz

参考代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pi pair<int,int>
#define Mp make_pair
const int inf = 0x3f3f3f3f;
const int N = 5e5+10;
#define ll long long
using namespace std;

ll a[N],sum[N],n,len;
double fun(int x)
{
    return (double)a[len]-((sum[x]+a[len])/(double)(x+1));
}
int main()
{
    ll x,l,r,lmid,rmid;
    double maxx;
    sum[0]=0;
    len=0;
    scanf("%lld",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&x);
        if(x==1)
        {
            len++;
            scanf("%lld",&a[len]);
            sum[len]=sum[len-1]+a[len];
        }
        else
        {
            r=len-1;
            l=1;
            while(r-l>2)
            {
                lmid=l+(r-l)/3;
                rmid=r-(r-l)/3;
                if(fun(lmid)>fun(rmid))
                    r=rmid;
                else if(fun(lmid)<fun(rmid))
                    l=lmid;
                else
                {
                    l=lmid;
                    r=rmid;
                }
            }
            maxx=0;
            for(int j=l; j<=r; j++)
                maxx=max(maxx,fun(j));
            printf("%.8lf\n",maxx);
        }
    }
    return 0;
}