CF 143 div2 C

C. To Add or Not to Add

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A piece of paper contains an array of n integers a1, a2, ..., an. Your task is to find a number that occurs the maximum number of times in this array.

However, before looking for such number, you are allowed to perform not more than k following operations — choose an arbitrary element from the array and add 1 to it. In other words, you are allowed to increase some array element by 1 no more than k times (you are allowed to increase the same element of the array multiple times).

Your task is to find the maximum number of occurrences of some number in the array after performing no more than k allowed operations. If there are several such numbers, your task is to find the minimum one.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105; 0 ≤ k ≤ 109) — the number of elements in the array and the number of operations you are allowed to perform, correspondingly.

The third line contains a sequence of n integers a1, a2, ..., an (|ai| ≤ 109) — the initial array. The numbers in the lines are separated by single spaces.

Output

In a single line print two numbers — the maximum number of occurrences of some number in the array after at most k allowed operations are performed, and the minimum number that reaches the given maximum. Separate the printed numbers by whitespaces.

Sample test(s)

input

5 3
6 3 4 0 2

output

3 4

input

3 4
5 5 5

output

3 5

input

5 3
3 1 2 2 1

output

4 2

Note

In the first sample your task is to increase the second element of the array once and increase the fifth element of the array twice. Thus, we get sequence 6, 4, 4, 0, 4, where number 4 occurs 3 times.

In the second sample you don't need to perform a single operation or increase each element by one. If we do nothing, we get array 5, 5, 5, if we increase each by one, we get 6, 6, 6. In both cases the maximum number of occurrences equals 3. So we should do nothing, as number 5 is less than number 6.

In the third sample we should increase the second array element once and the fifth element once. Thus, we get sequence 3, 2, 2, 2, 2, where number 2 occurs 4 times.

因为a的范围很大，而n的范围不大，所以可以考虑枚举最终哪个数出现的次数最多，而要使某个数x出现次数尽量多，就是每次先找比它小的，离它最近的那个数，把那个数加到x，但是对每个数都这样去找，是不行的，因为时间最坏情况是n^2。于是，考虑怎么优化。先排序是必须的，然后我是否能利用x前一个数的出现的最多次数直接得到x出现的最多次数呢？可以的，先考虑把x前一个数都变成x，如果当前的k还够，那就直接变，k不够，根据贪心的思想，我会把所有变成x前一个数的数里，那个最小的扔掉。这样复杂度就是2*n，O(n).最后，数据比较大，要用long long。

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn = 100000 + 5;

LL a[maxn];

int main(){
    int n,k;
    while(cin >> n >> k){
        for(int i = 0;i < n;i++) cin >> a[i];
        sort(a,a+n);
        int l = 0;
        LL ans = a[0],cnt = 1,ansx = 1;
        for(int i = 1;i < n;i++){
            if((a[i]-a[i-1])*cnt <= k){
                k -= (a[i]-a[i-1])*cnt;
                cnt++;
                if(cnt > ansx){
                    ansx = cnt;
                    ans = a[i];
                }
            }
            else{
                while((a[i]-a[i-1])*cnt > k){
                    k += a[i-1]-a[i-cnt];
                    cnt--;
                }
                k -= (a[i]-a[i-1])*cnt;
                cnt++;
                if(cnt > ansx){
                    ansx = cnt;
                    ans = a[i];
                }
            }
        }
        cout << ansx << ' ' << ans << endl;
    }
    return 0;
}