题目链接
bitset简直暴力，bitset+分块（暴力+暴力）
还是维护每个属性的前i个的位置，对于一组询问查询在各属性的upper_bound然后&起来就好了。
五维偏序模板。。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
const int maxn = 50005;
typedef long long ll;
const ll mod = 1e9+7;
int Case = 1;
int n, m, bk, kp;
int getbk(int x) {
    return (x-1)/bk+1;
}
struct node{
    int b[6];int id;
    bool operator<(const node s)const{
        return b[kp] < s.b[kp];
    }
}cc[maxn];
bitset<maxn>bit[6][500];
int a[6][maxn], id[6][maxn];
void solve() {
    kp = 1;
    memset(bit, 0, sizeof(bit));
    scanf("%d%d", &n, &m);
    bk = sqrt(n);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= 5; j++) {
            scanf("%d", &cc[i].b[j]);
        }
        cc[i].id = i;
    }
    for(int j = 1; j <= 5; j++) {
        sort(cc+1, cc+1+n);kp++;
        for(int i = 1; i <= n; i++) {
            a[j][i] = cc[i].b[j];id[j][i] = cc[i].id;
            int p = getbk(i);
            if(getbk(i-1) != p) bit[j][p] |= bit[j][p-1];
            bit[j][p][id[j][i]] = 1;
        }
    }
    int q, last = 0;
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        node tmp;
        for(int j = 1; j <= 5; j++) {
            scanf("%d", &tmp.b[j]);
            tmp.b[j] ^= last;
        }
        bitset<maxn>res;res.set();
        for(int j = 1; j <= 5; j++) {
            bitset<maxn>t;
            int x = tmp.b[j];
            int p = upper_bound(a[j]+1, a[j]+1+n, x)-a[j];
            if(p > n || a[j][p] > x) p--;
            int pk = getbk(p)-1;
            t = bit[j][pk];
            for(int sp = pk*bk+1; sp <= p; sp++) {
                t[id[j][sp]] = 1;
            }
            res &= t;
        }
        printf("%d\n", last = res.count());
    }
    return;
}
int main() {
    //g++ -std=c++11 -o2 1.cpp -o f && ./f < in.txt
    //ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt","w",stdout);
#endif
    scanf("%d", &Case);
    while(Case--) {
        solve();
        }
    return 0;
}