（线段树）CF813 Educational Codeforces Round 22 E

E. Army Creation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

首先计算每个编号的士兵前第k个与其类型相同的人的位置。对于这个士兵所在的查询区间[l,r]，如果上述前面那个人的位置小于l则这个士兵在[l,r]中同类型人的编号中一定小于k，即可以计算。

计算出上述的prek数组后，对其建立线段树以方便查询。线段树维护区间中的所有prek，维护成有序的形式，之后查询时直接使用lower_bound函数即可。

维护有序形式时，利用stl中的merge函数，将一个结点的两个子节点代表的区间合并即可。

merge函数的用法(第一个容器的起始位置，第一个容器的末位置，第二个容器的起始位置，第二个容器的末位置，存储合并完的容器的起始位置，（可有可无）cmp函数）

注意使用merge之前要对存储合并结果的容器进行resize操作，不然会报错。

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <vector>
#include <stack>
#define mp make_pair
//#define P make_pair
#define MIN(a,b) (a>b?b:a)
//#define MAX(a,b) (a>b?a:b)
typedef long long ll;
typedef unsigned long long ull;
const int MAX=1e5+5;
const int MAX_V=1e3+5;
const ll INF=4e18+5;
const double M=4e18;
using namespace std;
const int MOD=1e9+7;
typedef pair<ll,int> pii;
const double eps=0.000000001;
#define rank rankk
int la[MAX],prek[MAX];
vector<int>num[MAX<<2],cnt[MAX];
void build(int l,int r,int k)
{
    if(l==r)
    {
        num[k].push_back(prek[l]);
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
    num[k].resize(num[2*k].size()+num[2*k+1].size());
    merge(num[2*k].begin(),num[2*k].end(),num[2*k+1].begin(),num[2*k+1].end(),num[k].begin());
    return;
}
inline int query(int l,int r,int ql,int qr,int x,int k)
{
    if(ql<=l&&qr>=r)
        return lower_bound(num[k].begin(),num[k].end(),x)-num[k].begin();
    int mid=(l+r)/2;
    int re=0;
    if(ql<=mid)
        re+=query(l,mid,ql,qr,x,2*k);
    if(qr>mid)
        re+=query(mid+1,r,ql,qr,x,2*k+1);
    return re;
}
int n,k,tem;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&tem);
        cnt[tem].push_back(i);
        if(cnt[tem].size()>k)
            prek[i]=cnt[tem][cnt[tem].size()-k-1];
    }
    int m,an=0;
    build(1,n,1);
    scanf("%d",&m);
    while(m--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        x=(x+an)%n+1;
        y=(y+an)%n+1;
        if(y<x)
            swap(x,y);
        an=query(1,n,x,y,x,1);
        printf("%d
",an);
    }
    return 0;
}