/*
对区间的价值用树状数组维护
对颜色用线段树维护，即对每种颜色建立一棵线段树，动态开点保证空间，对于颜色i, 区间[l, r].sum 表示的是区间中颜色i的数量
*/
#include <bits/stdc++.h>
using namespace std;

#define ll long long
typedef vector<int> Vec;
const int N = 1e5 + 10, MAX = 1e5;

int n, q;

int v[N], c[N];

int T[N], tot; // T[i]:颜色为i的线段树的根
struct node{ 
    int ls, rs, sum;
}tr[N * 200];

void update(int& p, int loc, int k, int l = 1, int r = MAX){ // 对于该颜色更新点loc的价值+1
    if(!p) p = ++ tot;
    tr[p].sum += k;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(loc <= mid) update(tr[p].ls, loc, k, l, mid);
    else update(tr[p].rs, loc, k, mid + 1, r);
}

int query(int p, int ql, int qr, int l = 1, int r = MAX){ // 对于颜色i查询区间[ql, qr]该颜色的个数
    if(!p) return 0;
    if(ql <= l && r <= qr) return tr[p].sum;
    int mid = (l + r) >> 1, ans = 0;
    if(ql <= mid) ans += query(tr[p].ls, ql, qr, l, mid);
    if(qr > mid) ans += query(tr[p].rs, ql, qr, mid + 1, r);
    return ans; 
}

ll val[N]; // 树状数组维护单点修改和区间求和
int lowbit(int x){ return x & -x; }
void update(int r, int k){
    for(int i = r; i <= n; i += lowbit(i)) val[i] += k;
}
ll get_sum(int l, int r){
    ll ans = 0;
    for(int i = r; i; i -= lowbit(i)) ans += val[i];
    for(int i = l - 1; i; i -= lowbit(i)) ans -= val[i];
    return ans;
}

Vec ci; // 每次查询的颜色集合
bool check(int len, int l, int r){
    int sum = 0;
    for(auto x : ci){
        sum += query(T[x], l, r); // 对于所有集合中的颜色查询[l, r].sum
    }
    return sum == len;
}

void get_ans(){
    int x, k;
    cin >> x >> k;
    ci.clear();
    for(int i = 1, y; i <= k; i ++){
        cin >> y;
        ci.push_back(y);
    }
	
    sort(ci.begin(), ci.end());
    ci.erase(unique(ci.begin(), ci.end()), ci.end()); // 去重

    int l = 1, r = n;

    int Lr = x;
    while(l <= Lr){ // 二分找最远能到的左边界
        int mid = (l + Lr) >> 1;
        if(check(x - mid + 1, mid, x)) Lr = mid - 1;
        else l = mid + 1;
    }

    int Rl = x;
    while(Rl <= r){ // 二分找最远合法的右边界
        int mid = (Rl + r) >> 1;
        if(check(mid - x + 1, x, mid)) Rl = mid + 1;
        else r = mid - 1;
    }
    cout << get_sum(l, r) << "\n"; // 输出求和
}

void clear(){
     for(int i = 0; i <= tot; i ++) tr[i] = {0, 0, 0};
     for(int i = 1; i <= n; i ++) T[i] = val[i] = 0;
     tot = 0;
}
void solve(){
    cin >> n >> q;
    clear();
    for(int i = 1; i <= n; i ++){
        cin >> c[i];
        update(T[c[i]], i, 1);
    }
    for(int i = 1; i <= n; i ++){
        cin >> v[i];
        update(i, v[i]);
    }

    for(int i = 1; i <= q; i ++){
        int op, p, x;
        cin >> op;
        if(op == 1){ // 更改颜色
            cin >> p >> x;
            update(T[c[p]], p, -1);
            c[p] = x;
            update(T[c[p]], p, 1);
        }
        else if(op == 2){ // 更改价值
            cin >> p >> x;
            update(p, x - v[p]);
            v[p] = x;
        }
        else get_ans();
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    int t;
    cin >> t;
    while(t --){
    	solve();
    }
    return 0;
}