#include<iostream>

using namespace std;

const int N = 1e6 + 10;

typedef long long int ll;
const ll inf = 1e20;

ll n, q, tr[4 * N], add[4 * N], val[4 * N];

void push_up(int p)
{
	tr[p] = max(tr[2 * p], tr[2 * p + 1]);
}

void push_down(int p, int l, int r)
{
	if (val[p] <inf)//先处理改变为x的标记
	{
		tr[2 * p] = val[p];
		tr[2 * p + 1] = val[p];
		val[2 * p] = val[p];
		val[2 * p + 1] = val[p];
		add[2 * p] = add[2 * p + 1] = add[p] = 0;
		val[p] = inf;
	}
	if(add[p])//处理区间加法
	{
		int mid = (l + r) / 2;
		tr[2 * p] += add[p] ;
		tr[2 * p + 1] += add[p] ;
		add[2 * p] += add[p];
		add[2 * p + 1] += add[p];
		add[p] = 0;
	}
}

void build(int p, int l, int r)
{
	val[p] = inf;
	if (l == r)
	{
		cin >> tr[p];
		return;
	}

	int mid = (l + r) / 2;
	build(2*p, l, mid);
	build(2 * p + 1, mid + 1, r);
	push_up(p);
}

void change(int p, int l, int r, int x, int y, ll ad, ll vl)
{
	if (x <= l && r <= y)
	{
		if (vl <inf)//如果vl！=inf说明是操作1
			tr[p] = vl,add[p]=0,val[p]=vl;
		else//否则操作2
			tr[p] += ad,add[p] += ad;
		return;
	}
	push_down(p, l, r);
	int mid = (l + r) / 2;
	if (x <= mid)change(2*p, l, mid, x, y, ad, vl);
	if (y > mid)change(2*p+1, mid + 1, r, x, y, ad, vl);
	push_up(p);
}

ll find(int p, int l, int r, int x, int y)
{
	if (x <= l && r <= y)
	{
		return tr[p];
	}
	push_down(p, l, r);
	ll mid = (l + r) / 2, ans = -inf;
	if (x <= mid)ans =max(ans,find(2*p, l, mid, x, y));
	if (y > mid)ans =max(ans,find(2*p+1, mid + 1, r, x, y));
	return ans;
}

int main()
{
	cin >> n >> q;
	build(1, 1, n);
	ll op, l, r, x;
	while (q--)
	{
		cin >> op >> l >> r;
		if (op == 1)
		{
			cin >> x;
			change(1, 1, n, l, r, 0, x);
		}
		else if (op == 2)
		{
			cin >> x;
			change(1, 1, n, l, r, x,inf);
		}
		else
		{
			cout << find(1, 1, n, l, r)<<endl;
		}
	}
}