大概就是要多加一维time
然后按照(l的块，r的块，time)为关键字排序
转移区间修改还是按照莫队的方式（每个修改要记修改前后的状态）
然后玄学dalao告诉窝块大小设为(O(n^{frac{2}{3}}))最优
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cmath>
#define il inline
#define rg register
#define vd void
#define sta static
typedef long long ll;
il int gi(){
rg int x=0,f=1;rg char ch=getchar();
while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
const int maxn=10010;
int A[maxn];
int p[maxn],pa[maxn],pb[maxn];
struct ques{int l,r,id,t;}Q[maxn];
int ans[maxn];
int block[maxn];
bool operator <(const ques&a,const ques&b){
if(block[a.l]!=block[b.l])return block[a.l]<block[b.l];
if(block[a.r]!=block[b.r])return block[a.r]<block[b.r];
return a.t<b.t;
}
int tot[1000010],nowans;
il vd fuck(int a,int b){
if(tot[a]==1&&b==-1)--nowans;
if(tot[a]==0&&b==1)++nowans;
tot[a]+=b;
}
int main(){
#ifdef xzz
freopen("1903.in","r",stdin);
freopen("1903.out","w",stdout);
#endif
int n=gi(),m=gi(),k=0,q=0,blo=pow(n,0.67),a,b;
char o[2];
for(rg int i=1;i<=n;++i)A[i]=gi();
for(rg int i=1;i<=n;++i)block[i]=i/blo;
for(rg int i=1;i<=m;++i){
scanf("%s",o),a=gi(),b=gi();
if(o[0]=='Q')++q,Q[q].l=a,Q[q].r=b,Q[q].id=q,Q[q].t=k;
else if(o[0]=='R')++k,p[k]=a,pa[k]=A[a],A[a]=pb[k]=b;
}
std::sort(Q+1,Q+q+1);
int L=1,R=1,T=k;
tot[A[1]]=1;nowans=1;
for(rg int i=1;i<=q;++i){
while(T<Q[i].t){
++T,A[p[T]]=pb[T];
if(L<=p[T]&&p[T]<=R)fuck(pa[T],-1),fuck(pb[T],1);
}
while(T>Q[i].t){
A[p[T]]=pa[T];
if(L<=p[T]&&p[T]<=R)fuck(pb[T],-1),fuck(pa[T],1);
--T;
}
while(L>Q[i].l)--L,fuck(A[L],1);
while(R>Q[i].r)fuck(A[R],-1),--R;
while(R<Q[i].r)++R,fuck(A[R],1);
while(L<Q[i].l)fuck(A[L],-1),++L;
ans[Q[i].id]=nowans;
}
for(rg int i=1;i<=q;++i)printf("%d
",ans[i]);
return 0;
}