尺取+二进制 WiFi Password

Just days before the JCPC, your internet service went down. You decided to continue your training at the ACM club at your university. Sadly, you discovered that they have changed the WiFi password. On the router, the following question was mentioned, the answer is the WiFi password padded with zeros as needed.

A subarray [l, r] of an array A is defined as a sequence of consecutive elements A_l, A_l + 1, ..., A_r, the length of such subarray is r - l + 1. The bitwise OR of the subarray is defined as: A_l OR A_l + 1 OR ... OR A_r, where OR is the bitwise OR operation (check the notes for details).

Given an array A of n positive integers and an integer v, find the maximum length of a subarray such that the bitwise OR of its elements is less than or equal to v.

Input

The first line contains an integer T (1 ≤ T ≤ 128), where T is the number of test cases.

The first line of each test case contains two space-separated integers n and v (1 ≤ n ≤ 10⁵) (1 ≤ v ≤ 3 × 10⁵).

The second line contains n space-separated integers A₁, A₂, ..., A_n(1 ≤ A_i ≤ 2 × 10⁵), the elements of the array.

The sum of n overall test cases does not exceed 10⁶.

Output

For each test case, if no subarray meets the requirement, print 0. Otherwise, print the maximum length of a subarray that meets the requirement.

Example

Input

3
5 8
1 4 5 3 1
5 10
8 2 6 1 10
4 2
9 4 5 8

Output

5
3
0

Note

To get the value of x OR y, consider both numbers in binary (padded with zeros to make their lengths equal), apply the OR operation on the corresponding bits, and return the result into decimal form. For example, the result of 10 OR 17 = 01010 OR 10001 = 11011 = 27.

题目大意：从一个数组里找一段区间，把区间里的所有值or起来计算结果，求结果不超过v的最长区间。

 右端点每次往右移
大于v时更新左端点
更新时每次把数组上的对应数减一
数组用来记录二进制数位上的值

 记录第i位上有几个
然后去掉一个值的时候 把对应数位减一

 #include <cstdio>
#include<string.h>
int f[50];

int jisuan()
{
int ans=0;

int p=1;
for (int i=0;i<=30;i++)
{
if (f[i]) ans+=p;
  p*=2;
}

return ans;
}

int main() {
    int l,r,n,v,i,u,t,o,a[100005];
    
    scanf("%d",&t);
    
    for (u=0;u<t;u++)
    {
     scanf("%d%d",&n,&v);
     for (i=0;i<n;i++) scanf("%d",&a[i]);
    
     l=0;r=0;o=0;
     memset(f,0,sizeof(f));
     while (r<n)
     {
     while (jisuan()<=v)
{
if ((r-l)>o) o=r-l;
int k=0,q=a[r];
while (q>0)
{
f[k]+=q%2;
k++;
q/=2;
} 
r++;
if (r>n) break;
}
     while (jisuan()>v)
     {
     int k=0,q=a[l];
while (q>0)
{
f[k]-=q%2;
k++;
q/=2;
}
l++;
if (l>=r&&r>=n) break;
}
}
     printf("%d\n",o);
}
    
    return 0;
}