Codeforces Round #271 (Div. 2) E. Pillars 线段树优化dp

E. Pillars

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Marmot found a row with n pillars. The i-th pillar has the height of hi meters. Starting from one pillar i1, Marmot wants to jump on the pillars i2, ..., ik. (1 ≤ i1 < i2 < ... < ik ≤ n). From a pillar i Marmot can jump on a pillar j only if i < j and |hi - hj| ≥ d, where |x| is the absolute value of the number x.

Now Marmot is asking you find out a jump sequence with maximal length and print it.

Input

The first line contains two integers n and d (1 ≤ n ≤ 105, 0 ≤ d ≤ 109).

The second line contains n numbers h1, h2, ..., hn (1 ≤ hi ≤ 1015).

Output

The first line should contain one integer k, the maximal length of a jump sequence.

The second line should contain k integers i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), representing the pillars' indices from the maximal length jump sequence.

If there is more than one maximal length jump sequence, print any.

Examples

input

5 2
1 3 6 7 4

output

4
1 2 3 5 

input

10 3
2 1 3 6 9 11 7 3 20 18

output

6
1 4 6 7 8 9 

Note

In the first example Marmot chooses the pillars 1, 2, 3, 5 with the heights 1, 3, 6, 4. Another jump sequence of length 4 is 1, 2, 4, 5.

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-8
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=2e6+10,inf=1e9+10,mod=1e9+7;
const LL INF=1e18+10;

struct SGT
{
    int maxx[N<<2];
    void build(int l,int r,int pos)
    {
        memset(maxx,0,sizeof(maxx));
    }
    void update(int p,int c,int l,int r,int pos)
    {
        if(l==r)
        {
            maxx[pos]=c;
            return;
        }
        int mid=(l+r)>>1;
        if(p<=mid)update(p,c,l,mid,pos<<1);
        else update(p,c,mid+1,r,pos<<1|1);
        maxx[pos]=max(maxx[pos<<1],maxx[pos<<1|1]);
    }
    int query(int L,int R,int l,int r,int pos)
    {
        if(L<=l&&r<=R)return maxx[pos];
        int mid=(l+r)>>1;
        int ans=0;
        if(L<=mid)ans=max(ans,query(L,R,l,mid,pos<<1));
        if(R>mid)ans=max(ans,query(L,R,mid+1,r,pos<<1|1));
        return ans;
    }
}tree;
LL a[N],b[N];
int n;
LL k;
int big(LL x)
{
    int pos=lower_bound(b,b+2+n,x)-b;
    return pos;
}
int low(LL x)
{
    int pos=upper_bound(b,b+2+n,x)-b-1;
    return pos;
}
int dp[N];
vector<int>ans;
int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        b[i]=a[i];
    }
    sort(b+1,b+1+n);
    b[n+1]=INF;
    b[0]=-INF;
    for(int i=1;i<=n;i++)
    {
        LL pre=a[i]-k;
        LL nex=a[i]+k;
        int pos1=low(pre);
        int pos2=big(nex);
        //cout<<pre<<" "<<nex<<" "<<pos1<<" "<<pos2<<endl;
        int maxx=0;
        if(pos1>=1)maxx=max(maxx,tree.query(1,pos1,1,n,1));
        if(pos2<=n)maxx=max(maxx,tree.query(pos2,n,1,n,1));
        dp[i]=maxx+1;
        tree.update(low(a[i]),maxx+1,1,n,1);
    }
    int maxx=0,pre=-1;
    for(int i=1;i<=n;i++)
    {
        if(dp[i]>maxx)
        {
            maxx=dp[i];
            pre=i;
        }
    }
    ans.push_back(pre);
    for(int i=pre-1;i>=1;i--)
    {
        if(abs(a[i]-a[pre])>=k&&dp[pre]==dp[i]+1)
        {
            pre=i;
            ans.push_back(pre);
        }
    }
    printf("%d\n",maxx);
    for(int i=maxx-1;i>=0;i--)
        printf("%d ",ans[i]);
    return 0;
}