CodeForces - 748E (枚举+脑洞)

E. Santa Claus and Tangerines

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Santa Claus has n tangerines, and the i-th of them consists of exactly a_i slices. Santa Claus came to a school which has k pupils. Santa decided to treat them with tangerines.

However, there can be too few tangerines to present at least one tangerine to each pupil. So Santa decided to divide tangerines into parts so that no one will be offended. In order to do this, he can divide a tangerine or any existing part into two smaller equal parts. If the number of slices in the part he wants to split is odd, then one of the resulting parts will have one slice more than the other. It's forbidden to divide a part consisting of only one slice.

Santa Claus wants to present to everyone either a whole tangerine or exactly one part of it (that also means that everyone must get a positive number of slices). One or several tangerines or their parts may stay with Santa.

Let b_i be the number of slices the i-th pupil has in the end. Let Santa's joy be the minimum among all b_i's.

Your task is to find the maximum possible joy Santa can have after he treats everyone with tangerines (or their parts).

Input

The first line contains two positive integers n and k (1 ≤ n ≤ 10⁶, 1 ≤ k ≤ 2·10⁹) denoting the number of tangerines and the number of pupils, respectively.

The second line consists of n positive integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁷), where a_i stands for the number of slices the i-th tangerine consists of.

Output

If there's no way to present a tangerine or a part of tangerine to everyone, print -1. Otherwise, print the maximum possible joy that Santa can have.

Examples

Input

3 2
5 9 3

Output

5

Input

2 4
12 14

Output

6

Input

2 3
1 1

Output

-1

题意：n个橘子分给m个人，没个橘子有num[i]瓣，求获得最少的人最多获得多少瓣
题解:当num[i]的和小于m，输出-1.
否则，维护最大的答案，每次将最大的橘子分成两瓣，更新最大答案，不可分则退出

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 10000010
#define CLR(a,b) memset(a,b,sizeof(a))
long long num[maxn];
int main()
{
    int n,m;
    long long sum=;
    scanf("%d %d",&n,&m);
    for(int i=;i<=n;i++){
        int x;
        scanf("%d",&x);
        num[x]++;
        sum+=x;
    }

    if(sum < m){
        printf("-1\n");
        return ;
    }

    sum = ;
    int left=;
    for(int i=maxn;i>;i--){
        sum+=num[i];
        if(sum>=m){     //如果橘子数量大于m个小朋友，则最小的橘子为最小答案
            left=i;
            break;
        }
    }
    for(int i=maxn;i>;i--){
        if(i/<left)
            break;
        num[i/]+=num[i];
        num[i-i/]+=num[i];
        sum+=num[i];    //sum[i]已经包含一个num[i],＋num[i]相当于分成两瓣
        num[i]=;
        while(sum-num[left]>=m){
            sum-=num[left];
            left++;
        }
    }

    printf("%d\n",left);

    return ;
}