Codeforces Round #143 (Div. 2) C. To Add or Not to Add 胡搞

A piece of paper contains an array of n integers a₁, a₂, ..., a_n. 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, ..., a_n (|a_i| ≤ 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.

         题意：给你一个数列（好吧，又是数列= =），可对这个数列中的某些数加一个值，但加的总值不能超过k。求在这些操作下，使得数列中的某个数出现的次数最多。

      这个题纯靠yy，我们可以依次从数列中最小的数开始枚举，但复杂度为O(n^2)，铁定超时。但是可以优化之，大大降低复杂度。首先对数列排序，然后依次从左往右开始枚举，那么这里可以有一个状态转移：假设当前状态还剩余d可操作，满足数相等的最左端的数的位置为l，那么新加一个数num[i+1],只要比较(num[i+1]-num[i])*(i-l+1)与d的关系即可，来改变l的位置，那么就能很快求出下一状态。

      My code：

//STATUS:C++_AC_78MS_400KB 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL __int64
#define Max(x,y) ((x)>(y)?(x):(y))
#define Min(x,y) ((x)<(y)?(x):(y))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a))
const int MAX=100010,INF=200000000,MOD=1000000007;
const double esp=1e-6;
int cmp(const void *a,const void *b){
	return *(int*)a - *(int*)b;
}
int num[MAX];
int n;
LL k;

int main()
{
//	freopen("in.txt","r",stdin);
	int i,j,l,max_cou,max_num;
	LL d;
	while(~scanf("%d%I64d",&n,&k))
	{
		for(i=1;i<=n;i++)
			scanf("%d",&num[i]);

		qsort(num+1,n,sizeof(int),cmp);
		l=1;
		max_cou=1,max_num=num[1];
		for(i=2;i<=n;i++){
			d=(LL)(i-l)*(LL)(num[i]-num[i-1]);
			k-=d;
			while(k>0 && l>1 && num[i]-num[l-1]>=k){
				l--;
				k-=num[i]-num[l];
			}
			while(k<0 && lmax_cou){
				max_cou=i-l+1;
				max_num=num[i];
			}
		}
		
		printf("%d %d\n",max_cou,max_num);
	}
    return 0;
}