暴力分块，多次尝试仍90 TLE，求助

如题，尝试用分块模板过，但是不知为何开了O2只能拿90，哪怕是用了从ST表模板题题面复制的快读模板也过不了……为什么不用ST表？因为我ST表模板题是靠分块过的……用了O2之后#8 和#10 都能卡在500ms左右，但是#9 就是TLE，请各位dalao们伸出援手，用更多的常数优化卡过这道题，或者让这个*山一样的万年模板能用臭氧冲过去，在此先谢过各位了

以下是极度丑陋的代码

#include<bits/stdc++.h>
#define maxn 1000005 
using namespace std;

//devised from my P3372 Segment Tree I Template: only 50 points is expected this way
//First attempt with O2 gave 90; pumping ozone smashed the results to 0. 

long long n,m,len,a[maxn],val_max[maxn],val_min[maxn],be[maxn];

/*
void modify(long long from,long long to,long long ad){
    for(long long i=from;i<=min(to,be[from]*len);i++) a[i]+=ad,val[be[from]]+=ad;
    if(be[from]!=be[to]){
        for(long long i=(be[to]-1)*len+1;i<=to;i++) a[i]+=ad,val[be[to]]+=ad;
    }
    for(long long i=be[from]+1;i<=be[to]-1;i++) add[i]+=ad;
}
*/

inline int read(){ //This is from the ST chart template but it did NOT work sufficiently
    int x=0,f=1;char ch=getchar();
    while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}
    while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}

long long query_min(long long from,long long to){
    long long cnt=9223372036854775805;
    for(long long i=from;i<=min(to,be[from]*len);i++) cnt=min(cnt, a[i]);
    if(be[from]!=be[to]){
        for(long long i=(be[to]-1)*len+1;i<=to;i++) cnt=min(cnt, a[i]);
    }
    for(long long i=be[from]+1;i<=be[to]-1;i++) cnt=min(cnt, val_min[i]);
    return cnt;
}

long long query_max(long long from,long long to){
    long long cnt=-9223372036854775805;
    for(long long i=from;i<=min(to,be[from]*len);i++) cnt=max(cnt, a[i]);
    if(be[from]!=be[to]){
        for(long long i=(be[to]-1)*len+1;i<=to;i++) cnt=max(cnt, a[i]);
    }
    for(long long i=be[from]+1;i<=be[to]-1;i++) cnt=max(cnt, val_max[i]);
    return cnt;
}

int main(){
    n=read();m=read();
    len=sqrt(n);
    for(long long i=1;i<=n;i++) be[i]=(i-1)/len+1;
    for(long long i=1;i<=n;i++) {
        val_min[i]=9223372036854775805;
        val_max[i]=-9223372036854775805;
    }
    for(long long i=1;i<=n;i++){
        a[i]=read();
        val_min[be[i]]=min(val_min[be[i]],a[i]);
        val_max[be[i]]=max(val_max[be[i]],a[i]);
    }
    for(long long i=1;i<=n-m+1;i++){
        printf("%d ",query_min(i,i+m-1));
    }
    printf("\n");
    for(long long i=1;i<=n-m+1;i++){
        printf("%d ",query_max(i,i+m-1));
    }
}