被以前自己瞎YY的东西坑了T T...单调队列的确是可以维护这种操作的....

　　显然这题可以转化成维护不在车上的东西的最小值， 支持插入和删去最早出现的值，然后就可以用单调队列了T T

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           #define ll long long#define uint unsigned intusing namespace std;const int maxn=20000010, inf=1e9;int n, m, a, b, d, T, l, r, L, R, up, fir, firans;int p[maxn], h[maxn], q[maxn];uint ans_sum, cur_ans;    bool fly[maxn], car[maxn];namespace IO{    int c;    unsigned int seed;    unsigned int randnum(){        seed^=seed<<13;        seed^=seed>>17;        seed^=seed<<5;        return seed;    }    inline int read(int &x){scanf("%d",&x);return x;}    inline void init_case(int &m,int &a,int &b,int &d,int p[]){        scanf("%d%u%d%d%d%d",&m,&seed,&a,&b,&c,&d);        for(int i=1;i<=m;i++){            if(randnum()%c==0) p[i]=-1;            else p[i]=randnum()%b;        }    }    inline void update_ans(unsigned int &ans_sum,unsigned int cur_ans,int no){        const static unsigned int mod=998244353;        ans_sum^=(long long)no*(no+7)%mod*cur_ans%mod;    }}using IO::read;using IO::init_case;using IO::update_ans;inline void qpush(int x){    while(L<=R && q[R]>=x) R--;    q[++R]=x;}inline int min(int a, int b){
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