CodeForces813E:Army Creation (主席树---上一题的加强版)

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; ith 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 l_i 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 a₁, a₂, ... a_n (1 ≤ a_i ≤ 100000).

The third line contains one integer q (1 ≤ q ≤ 100000).

Then q lines follow. ith line contains two numbers x_i and y_i 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 l_i and r_i for the ith plan, you have to do the following:

1. l_i = ((x_i + last) mod n) + 1;
2. r_i = ((y_i + last) mod n) + 1;
3. 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

题意：给定N个数，以及Q个询问，每个询问给出L和R，现在问在这个区间选最多的数，使得每个数出现次数不能大于K。

思路：有了上一题的积累，这题就是一眼题了。上一题相当于的K=1。此题，我们任然算前缀的贡献，[1,i]区间建立在[1,i-1]基础上，如果前缀[1,i]里a[i]出现的次数大于等于K，我们删去最前面那一个的贡献1，加上i的贡献1，保证每个前缀不同数的贡献小于等于K。

#include<bits/stdc++.h>
using namespace std;
const int maxn=200010;
struct node{ 
    int val,l,r; 
    node() {} 
    node(int L,int R,int V):l(L),r(R),val(V){}
}s[maxn*20];
int rt[maxn],cnt,ans;
queue<int>q[maxn];
void insert(int &now,int pre,int L,int R,int pos,int val)
{
    s[now=++cnt]=node(s[pre].l,s[pre].r,s[pre].val+val);  //先假设都等 
    if(L==R) return ;
    int Mid=(L+R)>>1;
    if(pos<=Mid) insert(s[now].l,s[pre].l,L,Mid,pos,val); //这里再把不等的改掉 
    else insert(s[now].r,s[pre].r,Mid+1,R,pos,val);
}
int query(int now,int pos,int L,int R)
{
    if(L==R) return s[now].val;
    int Mid=(L+R)>>1;
    if(pos<=Mid) return query(s[now].l,pos,L,Mid)+s[s[now].r].val;
    return query(s[now].r,pos,Mid+1,R);
}
int main()
{
    int N,Q,K,x,y,i;
    scanf("%d%d",&N,&K);
    for(i=1;i<=N;i++){
        scanf("%d",&x);
        if(q[x].size()<K) insert(rt[i],rt[i-1],1,N,i,1);
        else {
            int last=q[x].front(); q[x].pop();
            insert(y,rt[i-1],1,N,last,-1);
            insert(rt[i],y,1,N,i,1);
        }
        q[x].push(i);
    }
    scanf("%d",&Q);
    while(Q--){
        scanf("%d%d",&x,&y);
        x=(x+ans)%N+1;
        y=(y+ans)%N+1;
        if(x>y) swap(x,y);
        ans=query(rt[y],x,1,N);
        printf("%d\n",ans);
    }
    return 0;
}