To Add or Not to Add CodeForces - 231C

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.

Examples

input

5 3
6 3 4 0 2

output

3 4

input

3 45 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.

题意：就是给你n个数，对每个数最多使用k次+1的操作，然后求出最多k次+1操作后这n个数中最小的众数，同时输出次数。

分析：一看到这道题，就是枚举n个可能的数，但是这样根据n和k的范围，肯定会超时。那么我们换个思路想，假设ans是最后的答案，数量是m，也就是说在小于ans（因为能够进行+1操作必定是小于ans的数）的所有数中，有m-1个数与ans差值的和小于k。那么要使m最大，必然要使差值和最小咯，那些占用k太多资源的数在某些情况下需要剔除。因此我们要首先对这个数列进行排序，那么对于每个数a[i]，我们用（i-j+1）表示a[i]的个数，j<=i。那么在这种情况下我们需要判断（i-j+1）*a[i]和sum+k，sum是i到j这i-j+1个数的和当(i-j+1)*a[i]>sum+k的时候，就是我们不能用k来补足“缺口”，这时候就要将最占k资源的数减去，那肯定就是a[j]了，因为它是离i最远的，那就是sum-=a[j++]，记得这里是要用while((i-j+1)*a[i]>sum+k)来判断的，因为我们要知道后边数是否满足判断条件。最后比较个数就可以了，注意题目是要输出最小的众数

代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
long long k;
//freopen("test.txt","r",stdin);
while(~scanf("%d%I64d",&n,&k)){
long long a[100005];
for(int i=1;i<=n;i++){
scanf("%I64d",&a[i]);
}
sort(a+1,a+1+n);
long long ans1=0,ans2=0;
long long j=1;
long long sum=0;
for(long long i=1;i<=n;i++){
sum+=a[i];
while((i-j+1LL)*a[i]-sum>k){
sum-=a[j++];
}
if(i-j+1>ans2){
ans2=i-j+1;
ans1=a[i];
}
}
printf("%I64d %I64d\n",ans2,ans1);
}
return 0;
}