To Add or Not to Add CodeForces - 231C 尺取

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.

Example

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
<
4000
pre>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.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll M=1e5+10;
ll n,k;
ll a[M];
int main()
{
while(~scanf("%lld%lld",&n,&k))
{
for(ll i=0;i<n;i++)
{
scanf("%lld",&a[i]);
}
sort(a,a+n);
ll ans,res;
ll i=0,j=0;
ll mac,num=0,sum=0,r;
for(i=0;i<n;i++)
{
mac=a[i];
sum+=a[i];
if(sum+k>=mac*(num+1))
{
num++;
r=mac;
}
else
{
sum-=a[i];
break;
}
}
res=num;ans=r;
while(i<n)
{
sum-=a[j];
num--;
while(i<n)
{
sum+=a[i];
mac=a[i];
if(sum+k>=mac*(num+1))
{
num++;
r=mac;
i++;
}
else
{
sum-=a[i];
break;
}
}
if(num>res)
{
res=num;
ans=r;
}
j++;
}
printf("%lld %lld\n",res,ans);
}
return 0;
}