# include <bits/stdc++.h>

# define ffor(i,name) \
	for (auto i = name.begin (); i != name.end (); ++ i)

# define iter vector <int> :: iterator

using namespace std;

typedef long long ll;

typedef pair <int, int> pii;

int n, m, op, x, lst, ans, l[1100005], r[1100005], a[1100005], w[1100005], sum[1100005], len, rt;

void pushup (int x) {

	sum[x] = sum[l[x]] + sum[r[x]] + 1;

	return ;

}

pii divide (int x, int k) {

	if (! x)
		return {0, 0};

	if (a[x] < k) {

		pii son = divide (r[x], k);

		r[x] = son.first;

		pushup (x);

		return {x, son.second};

	}

	pii son = divide (l[x], k);

	l[0] = son.second;

	pushup (x);

	return {son.first, x};

}

int merge (int x, int y) {

	if (! x || ! y)
		return x + y;

	if (w[x] < w[y]) {

		r[x] = merge (r[x], y);

		pushup (x);

		return x;

	}

	l[y] = merge (x, l[y]);

	pushup (y);

	return y;

}

void insert (int x, int i) {

	a[++ len] = x;

	w[len] = i;

	sum[len] = 1;

	pii t = divide (rt, x);

	rt = merge (merge (t.first, len), t.second);

	return ;

}

void del (int x) {

	pii L = divide (rt, x), R = divide (L.second, x + 1);

	R.first = merge (l[R.first], r[R.first]);

	rt = merge (L.first, merge (R.first, R.second));

	return ;

}

int find1 (int x) {

	pii t = divide (rt, x);

	int ans = sum[t.first] + 1;

	rt = merge (t.first, t.second);

	return ans;

}

int find2 (int x) {

	int now = rt;

	while (now)
		if (sum[l[now]] + 1 == x)
			return a[now];

		else if (x <= sum[l[now]])
			now = l[now];
		else
			x -= sum[l[now]] + 1, now = r[now];

	return -1;

}

int last (int x) {

	return find2 (find1 (x) - 1);

}

int next (int x) {

	return find2 (find1 (x + 1));

}

int main () {

	ios::sync_with_stdio (0);

	cin.tie (0);

	cout.tie (0);

	cin >> n >> m;

	for (int i = 1; i <= n; ++ i)
		cin >> x, insert (x, i);

	for (int i = 1; i <= m; ++ i) {

		cin >> op >> x;

		x ^= lst;

		if (op < 2)
			insert (x, i + n);
		else if (op < 3)
			del (x);
		else if (op < 4)
			lst = find1 (x), ans ^= lst;
		else if (op < 5)
			lst = find2 (x), ans ^= lst;
		else if (op < 6)
			lst = last (x), ans ^= lst;
		else
			lst = next (x), ans ^= lst;

	}

	cout << ans;

	return 0;

}