Three-level Laser (贪心+二分)

Three-level Laser

An atom of element X can exist in n distinct states with energies E₁ < E₂ < ... < E_n. 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 E_k - E_i energy to put the atom in the state k,
3. the atom emits a photon with useful energy E_k - E_j and changes its state to the state j,
4. the atom spontaneously changes its state to the state i, losing energy E_j - E_i,
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 E_k - E_i ≤ 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 E_k and E_i.

The second line contains a sequence of integers E₁, E₂, ..., E_n (1 ≤ E₁ < E₂... < E_n ≤ 10⁹). It is guaranteed that all E_i 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

Note

In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to .

In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to .

题目给出一个序列，从小到大的顺序，求,的最大值其中i<j<k并且Ek - Ei <= U

思路：每次我们都选取两个相邻的数字作为Ei和Ej，这样可以保证他们的差值尽量小，尽量接近于1，接下来我们只需要在这两个数之后枚举每个数不断更新最大值就行了，发现数据量1e5直接暴力相当于n^2,会超时，所以后面的数字Ek的选取可以使用二分，若满足Ek-Ei<=U就找右边更大的数否则找左边。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e5+10;
int main(){
    int n,U;
    int a[maxn];
    scanf("%d%d",&n,&U);
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
    }
    double ans = -1;
    for(int i = 2; i <= n-1; i++){
        if(a[i+1] - a[i-1] > U) continue;
        int l = i + 1,r = n;
        while(l <= r){
            int mid = l + r >> 1;
            if(a[mid] - a[i-1] <= U)
                l = mid + 1;
            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;
}