[codeforces813E]Army Creation

time limit per test : 2 seconds
memory limit per test : 256 megabytes

As you might remember from our previous rounds, Vova really likes computer games. Now he is playing a strategy game known as Rage of Empires.

In the game Vova can hire $n$ different warriors; $i$ th warrior has the type $a_i$. Vova wants to create a balanced army hiring some subset of warriors. An army is called balanced if for each type of warrior present in the game there are not more than $k$ warriors of this type in the army. Of course, Vova wants his army to be as large as possible.

To make things more complicated, Vova has to consider $q$ different plans of creating his army. ith plan allows him to hire only warriors whose numbers are not less than li and not greater than $r_i$.

Help Vova to determine the largest size of a balanced army for each plan.

Be aware that the plans are given in a modified way. See input section for details.

Input

The first line contains two integers $n$ and $k (1 ≤ n, k ≤ 100000)$.
The second line contains n integers $a_1, a_2, ... a_n (1 ≤ a_i ≤ 100000)$.
The third line contains one integer $q (1 ≤ q ≤ 100000)$.
Then $q$ lines follow. $i$ th line contains two numbers $x_i$ and yi which represent ith plan $(1 ≤ x_i, y_i ≤ n)$.

You have to keep track of the answer to the last plan (let’s call it last). In the beginning $last = 0$. Then to restore values of li and ri for the ith plan, you have to do the following:

$l_i = ((x_i + last) mod n) + 1$;
$r_i = ((y_i + last) mod n) + 1$;
If $l_i > r_i$, swap $l_i$ and $r_i$.

Output

Print q numbers. ith number must be equal to the maximum size of a balanced army when considering ith plan.

Example

Input

6 2
1 1 1 2 2 2
5
1 6
4 3
1 1
2 6
2 6

Output

2
4
1
3
2

Note

In the first example the real plans are:

1. 1 2
2. 1 6
3. 6 6
4. 2 4
5. 4 6

題意：
給定一個n，k,還有一個長度爲n的數組$a_1,a_2,a_3,...,a_n$，$a_i$表示第$i$個士兵的種類。一個合法的隊伍是保證每一種類的士兵個數不會超過$k$，有$q$組詢問，每組詢問一個區間，求區間內元素可以組成的最大合法隊伍是多大。強制在線。

題解：
對於每個$i$處理出從他往前數第$k+1$個$a[i]$的位置，記錄爲$pa[i]$，然後對於求答案我們只需要求$[l,r]$中，$pa[i]<l$的數量即爲答案，這個可以用主席樹解決。

#include<bits/stdc++.h>
#define LiangJiaJun main
using namespace std;
int n,k,q,m,a[100004];
int pa[100004],mk[100004];
int rt[100004];
vector<int>vec[100004];
vector<int>v;
int tot;
struct prt{
    int l,r,w;
}T[100004*20];
int getid(int x){
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
void build(int &k,int l,int r){
     k=++tot;
     T[k].w=0;
     if(l==r)return ;
     int mid=(l+r)>>1;
     build(T[k].l,l,mid);
     build(T[k].r,mid+1,r);
}
void update(int l,int r,int &now,int last,int k){
     T[++tot]=T[last];
     now=tot;
     T[now].w++;
     if(l==r)return ;
     int mid=(l+r)>>1;
     if(k<=mid)update(l,mid,T[now].l,T[last].l,k);
     else update(mid+1,r,T[now].r,T[last].r,k);
}
int query(int l,int r,int x,int y,int k){
    if(l==r)return T[y].w-T[x].w;
    int mid=(l+r)>>1;
    if(k<=mid)return query(l,mid,T[x].l,T[y].l,k);
    else{
        int ret=0;
        ret+=T[T[y].l].w-T[T[x].l].w;
        ret+=query(mid+1,r,T[x].r,T[y].r,k);
        return ret;
    }
}
int LiangJiaJun(){
    tot=0;
    memset(rt,0,sizeof(rt));
    for(int i=0;i<=100000;i++)vec[i].clear();
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        vec[a[i]].push_back(i);
        mk[i]=vec[a[i]].size()-1;
    }
    for(int i=1;i<=n;i++){
        if(mk[i]>=k){
            pa[i]=vec[a[i]][mk[i]-k];
        }
        else{
            pa[i]=0;
        }
        v.push_back(pa[i]);
    }
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
    build(rt[0],1,n);
    for(int i=1;i<=n;i++){
        update(1,n,rt[i],rt[i-1],getid(pa[i]));
    }
    scanf("%d",&q);
    int last=0;
    while(q--){
        int l,r,k;
        scanf("%d%d",&l,&r);
        l=((l+last)%n+1);
        r=((r+last)%n+1);
        if(l>r)swap(l,r);
        k=upper_bound(v.begin(),v.end(),l-1)-v.begin();
        last=query(1,n,rt[l-1],rt[r],k);
        printf("%d\n",last);
    }
    return 0;
}