題目大意：

　　古今序列者，長皆​矣.今有一序列，其長亦若此，名之曰​.世人皆知其連續子序列之數為​矣.現有一士，欲取之最大值於各連續子序列也.今用此​值，構建新序列.序列,操作之本源也.於是生操作幾許.予君一數​，試問新序列中大於/小於/等於k之數有幾何？

solution：

　　調了好久好久好久的bug(淚).此題不難想到線段樹維護.首先將序列離散，然後開兩棵線段樹維護每個位置​左右小於等於​的連續序列長​,為防止重複,左閉右開即可.不難發現,該節點對新序列的貢獻為​,然後對於查詢操作,再開一棵線段樹維新序列即可.

code:

#include<iostream>
#include
 
<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<map>
#define R register
#define next kdjadskfj
#define debug puts("mlg")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned 
 
long long ull;
inline ll read();
inline void write(ll x);
inline void writeln(ll x);
inline void writesp(ll x);
ll n,Q;
ll a[610000],b[620000];
ll q;
ll dat[6100000];
ll type[610000],que[610000];
ll Dat[6100000],used[620000];
ll Count[620000][2];
inline void Update(ll p,ll l,ll r,ll k){
    if(l==r){Dat[p]=used[k];return;}
    ll mid
 
=l+r>>1;
    if(k<=mid) Update(p<<1,l,mid,k);
    else Update(p<<1|1,mid+1,r,k);
    Dat[p]=max(Dat[p<<1],Dat[p<<1|1]);
}

inline ll Query(ll p,ll l,ll r,ll u,ll v){
    if(!Dat[p]||(u<=l&&r<=v)) return Dat[p];
    ll mid=l+r>>1,Ans=0;
    if(u<=mid) Ans=max(Ans,Query(p<<1,l,mid,u,v));
    if(v>mid) Ans=max(Ans,Query(p<<1|1,mid+1,r,u,v));
    return Ans;
}

inline void update(ll p,ll l,ll r,ll k,ll val){
    if(l==r){dat[p]+=val;return;}
    ll mid=l+r>>1;
    if(k<=mid) update(p<<1,l,mid,k,val);
    else update(p<<1|1,mid+1,r,k,val);
    dat[p]=dat[p<<1]+dat[p<<1|1];
}

inline ll query(ll p,ll l,ll r,ll u,ll v){
    if(!dat[p]||(u<=l&&r<=v)) return dat[p];
    ll mid=l+r>>1,Ans=0;
    if(u<=mid) Ans+=query(p<<1,l,mid,u,v);
    if(v>mid) Ans+=query(p<<1|1,mid+1,r,u,v);
    return Ans;
}

inline void solve2(){
    for(R ll i=1;i<=n;i++){
        Count[i][0]=i-Query(1,1,q,a[i],q)-1;
        used[a[i]]=i;
        Update(1,1,q,a[i]);
    }
        memset(used,0,sizeof used);
        memset(Dat,0,sizeof Dat);
    for(R ll i=n;i>=1;i--){
        Count[i][1]=(n-i+1)-Query(1,1,q,a[i]+1,q)-1;
        used[a[i]]=n-i+1;
        Update(1,1,q,a[i]);
    }
    for(R ll i=1;i<=n;i++){
        update(1,1,q,a[i],(Count[i][0]+1)*(Count[i][1]+1));
    }
}

inline void solve1(){
    for(R ll i=1;i<=n;i++){
        ll maxn=0;
        for(R ll j=i;j<=n;j++){
            maxn=max(maxn,a[j]);
            update(1,1,q,maxn,1);
        }
    }
}

inline void work(){
    for(R ll i=1;i<=Q;i++){
        if(type[i]==1){
            if(que[i]==q) writeln(0);
            else writeln(query(1,1,q,que[i]+1,q));
            continue;    
        }
        if(type[i]==2){
            writeln(query(1,1,q,que[i],que[i]));
            continue;
        }
        if(type[i]==3){
            if(que[i]==1) writeln(0);
            else writeln(query(1,1,q,1,que[i]-1));
            continue;
        }
    }
}

int main(){
    freopen("jxthree.in","r",stdin);
    freopen("jxthree.out","w",stdout);
    n=read();Q=read();
    for(R ll i=1;i<=n;i++) a[i]=b[i]=read();
    for(R ll i=1;i<=Q;i++){
        char wn=getchar();
        while(wn!='>'&&wn!='<'&&wn!='=') wn=getchar();
        type[i]=((wn=='>')?1:((wn=='=')?2:3));
        b[n+i]=que[i]=read();
    }
    sort(b+1,b+n+Q+1);
    q=unique(b+1,b+Q+n+1)-b-1;
    for(R ll i=1;i<=n;i++){
        a[i]=lower_bound(b+1,b+q+1,a[i])-b;
    }
    for(R ll i=1;i<=Q;i++){
        que[i]=lower_bound(b+1,b+q+1,que[i])-b;
    }
    solve2();
    work();
}
inline ll read(){ll x=0,t=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') t=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*t;}
inline void write(ll x){if(x<0){putchar('-');x=-x;}if(x<=9){putchar(x+'0');return;}write(x/10);putchar(x%10+'0');}
inline void writesp(ll x){write(x);putchar(' ');}
inline void writeln(ll x){write(x);putchar('\n');}