// Om Namah Shivaya
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a, b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a, b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
inspired by some submissions from:
https://github.com/mostafa-saad/MyCompetitiveProgramming/blob/master/Olympiad/COI/COI-19-tenis.txt
the winners are on some pref of court 1
why? because if i cant win and j can win, pos1(i) < pos2(j), then i can win j and hence, can also win the tournament, which is a contradiction
same observation can be extended to all courts
winners lie on some pref of all courts
find the smallest k s.t the multisets a[1..k], b[1..k], c[1..k] are all equal, check if min_pos(i) <= k
let's say our current set initially equals a[1..k]
when adding guys from b[1..k] and c[1..k], no extra element should be added to the set, otherwise our condition is violated
rephrasing the above condition:
set_union(a[1..k],b[1..k],c[1..k]).size() = k, then ans is yes
verbally,
the #of distinct elements in the array {a[1..k]+b[1..k]+c[1..k]} = k
find the smallest k that satisfies this condition
to find the #of distinct elements in the array formed for some k, just look at the min_pos of each player in all 3 courts combined
add 1 at this min_pos for each player
find the first pos where sum(1..k)-k = 0
we know that sum(1..k)-k >= 0, so just find the min val and its leftmost occurrence
we dont have to find k, just check if i is winning or not, so just check if no zero lies on [1..min_pos(i)-1] i.e min(1..min_pos(i)-1) > 0
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
template<typename T>
struct lazysegtree {
/*=======================================================*/
struct data {
int a;
};
struct lazy {
int a;
};
data d_neutral = {inf1};
lazy l_neutral = {0};
void merge(data &curr, data &left, data &right) {
curr.a = min(left.a,right.a);
}
void create(int x, int lx, int rx, T v) {
}
void modify(int x, int lx, int rx, T v) {
lz[x].a = v;
}
void propagate(int x, int lx, int rx) {
int v = lz[x].a;
if(!v) return;
tr[x].a += v;
if(rx-lx > 1){
lz[2*x+1].a += v;
lz[2*x+2].a += v;
}
lz[x] = l_neutral;
}
/*=======================================================*/
int siz = 1;
vector<data> tr;
vector<lazy> lz;
lazysegtree() {
}
lazysegtree(int n) {
while (siz < n) siz *= 2;
tr.assign(2 * siz, d_neutral);
lz.assign(2 * siz, l_neutral);
}
void build(vector<T> &a, int n, int x, int lx, int rx) {
if (rx - lx == 1) {
if (lx < n) {
create(x, lx, rx, a[lx]);
}
return;
}
int mid = (lx + rx) / 2;
build(a, n, 2 * x + 1, lx, mid);
build(a, n, 2 * x + 2, mid, rx);
merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
}
void build(vector<T> &a, int n) {
build(a, n, 0, 0, siz);
}
void rupd(int l, int r, T v, int x, int lx, int rx) {
propagate(x, lx, rx);
if (lx >= r or rx <= l) return;
if (lx >= l and rx <= r) {
modify(x, lx, rx, v);
propagate(x, lx, rx);
return;
}
int mid = (lx + rx) / 2;
rupd(l, r, v, 2 * x + 1, lx, mid);
rupd(l, r, v, 2 * x + 2, mid, rx);
merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
}
void rupd(int l, int r, T v) {
rupd(l, r + 1, v, 0, 0, siz);
}
data query(int l, int r, int x, int lx, int rx) {
propagate(x, lx, rx);
if (lx >= r or rx <= l) return d_neutral;
if (lx >= l and rx <= r) return tr[x];
int mid = (lx + rx) / 2;
data curr;
data left = query(l, r, 2 * x + 1, lx, mid);
data right = query(l, r, 2 * x + 2, mid, rx);
merge(curr, left, right);
return curr;
}
data query(int l, int r) {
return query(l, r + 1, 0, 0, siz);
}
};
int a[5][N], pos[5][N];
void solve(int test_case)
{
int n,q; cin >> n >> q;
vector<int> mn_pos(n+5,inf1);
rep1(p,3){
rep1(i,n) cin >> a[p][i];
rep1(i,n){
int x = a[p][i];
pos[p][x] = i;
amin(mn_pos[x],i);
}
}
lazysegtree<int> st(n+5);
rep1(i,n){
st.rupd(i,i,-inf1-i);
}
auto upd = [&](int x, int add){
mn_pos[x] = inf1;
rep1(p,3){
amin(mn_pos[x],pos[p][x]);
}
st.rupd(mn_pos[x],n,add);
};
rep1(i,n){
upd(i,1);
}
while(q--){
int t; cin >> t;
if(t == 1){
int i; cin >> i;
if(st.query(1,mn_pos[i]-1).a > 0){
cout << "DA" << endl;
}
else{
cout << "NE" << endl;
}
}
else{
int p,x,y; cin >> p >> x >> y;
int i = pos[p][x], j = pos[p][y];
upd(x,-1);
upd(y,-1);
swap(a[p][i],a[p][j]);
swap(pos[p][x],pos[p][y]);
upd(x,1);
upd(y,1);
}
}
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
