二分答案找间距x，在栽树过程中每（x+1）棵树就必须栽至少一棵，类比传球游戏，用单调队列优化dp。

直接二分答案然后dp去check

线段树优化dp红红火火恍恍惚惚或或或或或或或或或或或或或或或或或

时间复杂度 $O(nlog^2n)$

#include<iostream>
#include<cstring>
#include<cstdio>
#define N 50005
#define int long long
using namespace std;
bool Test_MLE_start;
int T=1,n,m,ans=0;
int a[N],dp[N];
struct tree{
	int l,r,data;
}t[N<<2];
inline int reads(){
	char c=getchar();
	int sum=0,f=1;
	while(!isdigit(c)){
		if(c=='-') f=-1;
		c=getchar();
	}
	while(isdigit(c)){
		sum=(sum<<3)+(sum<<1)+(c^'0');
		c=getchar();
	}
	return sum*f;
}
inline void files(){
	freopen("E.in","r",stdin);
//	freopen("std.out","w",stdout);
}
inline void clr(){
//	Don't forget!

}
void pushup(int p){
	int x=p<<1,y=p<<1|1;
	t[p].data=min(t[x].data,t[y].data);
}
void builds(int p,int l,int r){
	t[p].l=l,t[p].r=r;
	if(l==r) return;
	int mid=(l+r)>>1;
	int x=p<<1,y=p<<1|1;
	builds(x,l,mid),builds(y,mid+1,r);
	pushup(p);
}
void changes(int p,int l,int r,int d){
	if(l<=t[p].l&&t[p].r<=r){
		t[p].data=d;
		return;
	}
	int mid=(t[p].l+t[p].r)>>1;
	int x=p<<1,y=p<<1|1;
	if(l<=mid) changes(x,l,r,d);
	if(r>mid) changes(y,l,r,d);
	pushup(p);
}
int asks(int p,int l,int r){
	if(l<=t[p].l&&t[p].r<=r) return t[p].data;
	int mid=(t[p].l+t[p].r)>>1;
	int x=p<<1,y=p<<1|1;
	int ans=2e9;
	if(l<=mid) ans=min(ans,asks(x,l,r));
	if(r>mid) ans=min(ans,asks(y,l,r));
	pushup(p);
	return ans;
}
bool check(int k){
	memset(dp,0,sizeof(dp));
	builds(1,0,n+1);
	for(int i=1;i<=n+1;i++){
		int x=asks(1,max(0ll,i-k-1),i-1);
		dp[i]=x+a[i];
		changes(1,i,i,dp[i]);
	}
//	cout<<k<<":";
//	for(int i=1;i<=n+1;i++) cout<<dp[i]<<" ";
//	puts("");
	return dp[n+1]<=m;
}
bool Test_MLE_end;
signed main(){
//	printf("%lf Mb\n",(&Test_MLE_end-&Test_MLE_start-1)/1024.0/1024.0);
//	files();
//	T=reads();
	while(T--){
		clr();
		n=reads(),m=reads();
		for(int i=1;i<=n;i++) a[i]=reads();
		int L=1,R=n;
		while(L<=R){
			int mid=(L+R)>>1;
			if(check(mid)){
				ans=mid;
				R=mid-1;
			}
			else L=mid+1;
		}
		printf("%lld\n",ans);
	}
	return 0;
}