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：
1 ：往集合里面加一个数，这个数比集合里所有元素都大
2：取一个子集使得 （子集中最大数max - 子集所有数的平均值)最大

题解：
先证该函数为凹函数，凹函数就跟一个v似的，有个凹点，两边都是凸起的，该题求解模式显然符合凹函数，凹函数类型可用三分思想求解

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
#define ll long long
ll a[maxn];
ll sum[maxn];
int cnt;
//三分思想
double rax(int x)
{
    return (double)a[cnt] - ((sum[x]+a[cnt])/(double)(x+1));
}
int main()
{
    int q;
    scanf("%d",&q);
    cnt = 0;
    for(int i=0; i<q; i++)
    {
        int x;
        scanf("%d",&x);
        if(x == 1)
        {
            cnt++;
            scanf("%lld",&a[cnt]);
            sum[cnt] = sum[cnt-1] + a[cnt];
        }
        else
        {
            int l = 1;
            int r = cnt;
            while(l<r-1)
            {
                int mid  = (l+r)/2;
                int mmid = (mid+r)/2;
                if( rax(mid) >= rax(mmid) )
                    r = mmid;
                else
                    l = mid;
            }
            double ans = 0;
            for(int i=l; i<=r; i++)
            {
                ans = max(ans, judge(i));
            }
            printf("%.10lf\n",ans);
        }
    }
}