VK Cup 2018 Round 2: B. Three-level Laser（二分）

C. Three-level Laser

time limit per test  1 second

memory limit per test  256 megabytes

input  standard input

output  standard output

An atom of element X can exist in n distinct states with energies E1 < E2 < ... < En. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.

Three distinct states i, j and k are selected, where i < j < k. After that the following process happens:

1. initially the atom is in the state i,
2. we spend Ek - Ei energy to put the atom in the state k,
3. the atom emits a photon with useful energy Ek - Ej and changes its state to the state j,
4. the atom spontaneously changes its state to the state i, losing energy Ej - Ei,
5. the process repeats from step 1.

Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.

Due to some limitations, Arkady can only choose such three states that Ek - Ei ≤ U.

Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.

Input

The first line contains two integers n and U (3 ≤ n ≤ 105, 1 ≤ U ≤ 109) — the number of states and the maximum possible difference between Ek and Ei.

The second line contains a sequence of integers E1, E2, ..., En (1 ≤ E1 < E2... < En ≤ 109). It is guaranteed that all Ei are given in increasing order.

Output

If it is not possible to choose three states that satisfy all constraints, print -1.

Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10 - 9.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

Examples

input

4 4
1 3 5 7

output

0.5

input

10 8
10 13 15 16 17 19 20 22 24 25

output

0.875

input

3 1
2 5 10

output

-1

题意：给你一个绝对递增的序列和一个数字U，选出三个数a[i], a[j], a[k]满足

①i<j<k；②a[k]-a[i]≤U；③(a[k]-a[j])/(a[k]-a[i])尽可能大

求③的最大值

因为只有a[k]-a[i]≤U的限制，而且输入的序列保证绝对递增

一个很容易想到的贪心就是a[i]和a[j]一定是挨着的，即i+1=j，然后这个k离它们越远越好

所以这题只要暴力a[j]，然后二分一下求出满足a[k]-a[i]≤U最大的a[k]，中间算出的最大值就是答案

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[100005];
int main(void)
{
	double ans;
	int n, i, U, l, r, mid;
	scanf("%d%d", &n, &U);
	for(i=1;i<=n;i++)
		scanf("%d", &a[i]);
	ans = -1;
	for(i=2;i<=n-1;i++)
	{
		l = i+1, r = n;
		if(a[i+1]-a[i-1]>U)
			continue;
		while(l<r)
		{
			mid = (l+r+1)/2;
			if(a[mid]-a[i-1]<=U)
				l = mid;
			else
				r = mid-1;
		}
		ans = max(ans, 1.0*(a[r]-a[i])/(a[r]-a[i-1]));
	}
	if(ans<-0.5)
		printf("-1\n");
	else
		printf("%.10f\n", ans);
	return 0;
}