Nikita and Order Statistics

Nikita likes tasks on order statistics, for example, he can easily find the k-th number in increasing order on a segment of an array. But now Nikita wonders how many segments of an array there are such that a given number x is the k-th number in increasing order on this segment. In other words, you should find the number of segments of a given array such that there are exactly k numbers of this segment which are less than x

.

Nikita wants to get answer for this question for each k

from 0 to n, where n

is the size of the array.

Input

The first line contains two integers n

and x (1≤n≤2⋅105,−109≤x≤109)

.

The second line contains n

integers a1,a2,…,an (−109≤ai≤109)

 — the given array.

Output

Print n+1

integers, where the i-th number is the answer for Nikita's question for k=i−1

.

Examples

Input

5 3
1 2 3 4 5

Output

6 5 4 0 0 0 

Input

2 6
-5 9

Output

1 2 0 

Input

6 99
-1 -1 -1 -1 -1 -1

Output

0 6 5 4 3 2 1 

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把所有小于 x 的数记为 1，剩下的数记为 0，则题目要求的是有多少个区间有 k 个 1。对这个 01 数组求前缀和，设为 s，则题目要求的是有多少个 (l,r)(l≤r) 满足 sr−sl−1=k

。

现在相当于有一个长度为 n+1

的数组 s，满足 si≤si+1≤si+1，问两个元素的差的取值的情况数。设多项式 f 的 xi 项的系数等于 sk=i 的 k 的个数，设多项式 g 的 x−i 项的系数等于 sk=i 的 k 的个数。对 f 和 g 做卷积，则积的 xk(k>0) 的系数就是对应的答案。当 k=0 时，首先要减去自己卷上自己的 n+1 次，然后再除以二，就是对应的答案。

#include"cstdio"
#include"iostream"
#include"cmath"
#include"cstring"
#include"algorithm"
using namespace std;
typedef long long ll;
int n,m;
const int nn=2e6+20;
const double pi = acos(-1);
struct complex {
  double x,y;
  complex(double xx=0,double yy=0){x=xx,y=yy;}
};

complex operator-(complex a,complex b) {return complex(a.x-b.x,a.y-b.y);}
complex operator+(complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
complex operator*(complex a,complex b)
{return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}

string s1,s2;
int n2; int k;
complex a[2000000],b[2000000];
int limit=1,l;
ll num[2000000];
int sum[2000000];
int r[2000000];

inline void fft(complex *t,int type)
{
   for (int i=0;i<limit;i++)if (i<r[i])swap(t[i],t[r[i]]);
   for (int mid=1;mid<limit;mid<<=1)
   {
     complex wn=complex(cos(pi/mid),type*sin(pi/mid));
     for (int R=mid<<1,j=0;j<limit;j+=R)
     {
        complex w=complex(1,0);
        for (int k=0;k<mid;k++,w=w*wn)
        {
           complex x=t[j+k],y=w*t[j+mid+k];
           t[j+k]=x+y;
           t[j+k+mid]=x-y;
        }
     }

   }
}
int t[300000];

int main()
{

   cin>>n>>k;
   for(int i=1;i<=n;i++)scanf("%d",&t[i]),t[i]=t[i-1]+(t[i]<k);
   for (int i=0;i<=n;i++)a[n+t[i]].x++,b[n-t[i]].x++;
   int len = 3*n;
   while (limit<=len)limit<<=1,l++;
   for (int i=0;i<limit;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
   fft(a,1); fft(b,1);
   for (int i=0;i<=limit;i++)a[i]=a[i]*b[i];
   fft(a,-1);
   for (int i=0;i<=len;i++)num[i]=(ll)(a[i].x/limit+0.5);
  // cout<<endl;
  // for (int i=1;i<=len;i++)cout<<num[i]<<" ";
  // cout<<"**************"<<endl;
   ll ans = num[n*2];
   ans=(ans-n-1)>>1;
   cout<<ans<<" ";
   for (int i=1;i<=n;i++)printf("%lld ",num[i+n+n]);


}