This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define all(v) v.begin(),v.end()
#define pb push_back
#define sz size()
#define ft first
#define sd second
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef unsigned long long ull;
const int N = 3e5 + 5;
const ll M = 1e8;
const ll inf = 1e3;
const ll mod = 1e9;
const double Pi = acos(-1);
ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res *= x;x *= x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); }
bool odd(ll n) { return (n % 2 == 1); }
bool even(ll n) { return (n % 2 == 0); }
int n, t, last, cur;
multiset<pair<int,pair<int, int>>> st;
pair<int,int> seg[N];
const void solve(/*Armashka*/) {
cin >> n >> t;
while (n --) {
int type; cin >> type;
if (type == 1) {
int a, b;
cin >> a >> b;
int l = a ^ (t * last), r = b ^ (t * last);
if (l > r) swap(l, r);
st.insert({cur + 1, {l, r}});
seg[++ cur] = {l, r};
} else if (type == 2) {
int id; cin >> id;
st.erase({id, seg[id]});
seg[id] = {0, 0};
} else {
int a, b, k, ans = 0;
cin >> a >> b >> k;
int l = a ^ (t * last), r = b ^ (t * last);
if (l > r) swap(l, r);
for (auto cur : st) {
int x = cur.sd.ft, y = cur.sd.sd;
if (max(x, l) <= min(y, r)) {
int cnt = min(y, r) - max(x, l) + 1;
if (cnt >= k) ++ ans;
}
}
cout << ans << "\n";
last = ans;
}
}
}
signed main() {
fast;
//freopen("divide.in", "r", stdin);
//freopen("divide.out", "w", stdout);
int tt = 1;
// cin >> tt;
while (tt --) {
solve();
}
}
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1
*/
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