#include<bits/stdc++.h>
#define lson (nw << 1)
#define rson (nw << 1 | 1)
using namespace std;
typedef long long ll;
inline ll read() {
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ '0');
        ch = getchar();
    }
    return x * f;
}
const int N = 2e5 + 10;
struct Op {
	int x, y, t; ll a, b; //x个栈，y的时间轴 
	Op(){}
	Op(int xx, int yy, int tt, ll aa, ll bb) {
		x = xx; y = yy; t = tt; a = aa; b = bb;
	}
	//t是type,  a, b是参数 
}a[N]; int tot, n, m, all; ll ans[N];
//!!!!!!注意longlong   线段树1~m
bool cmp(Op a, Op b) {
	if (a.x != b.x) return a.x < b.x;
	return a.t < b.t;
} 
struct Seg {
	int l, r; ll a, b, sum; //右括号， 左括号数， 以及所有数的和 
}tr[N*4];
ll merge(ll k, int nw) { //在x里面选择k个左括号 	
	if (!k) return 0;
	if (tr[nw].l == tr[nw].r) {
		if (!tr[nw].b) return 0;
		return min(tr[nw].b, k) * (tr[nw].sum / tr[nw].b);
	} ll lcnt = max(0ll, tr[lson].b - tr[rson].a);
	if (lcnt >= k) return merge(k, lson);
	else return tr[nw].sum - tr[rson].sum + merge(k - lcnt, rson); 
	//tr[nw].sum - tr[rson].sum 就是合并以后左边的sum,直接左边的sum是没有合并右边的 
	//如果不可减， 可以考虑再维护一个比如这题的：右边合并以后的左边sum, 或者楼房重建里的算上左边的max的时候右子树的sum 
}
void pushup(int nw) {
	if (tr[lson].b < tr[rson].a) {
		tr[nw].a = tr[lson].a + tr[rson].a - tr[lson].b; tr[nw].b = tr[rson].b;
		tr[nw].sum = tr[rson].sum;
	} else {
		tr[nw].a = tr[lson].a; tr[nw].b = tr[rson].b + tr[lson].b - tr[rson].a;
		tr[nw].sum = tr[rson].sum + merge(tr[lson].b - tr[rson].a, lson);
	} 
}
void build(int nw, int l, int r) {
	tr[nw].l = l; tr[nw].r = r;
	if (l == r) return;
	int mid = (l + r) >> 1; 
	build(lson, l, mid); build(rson, mid + 1, r);
}
void update(int nw, int x, ll a, ll b) { //a是左边的，也就是右括号， b是右边的，是左括号!!!!!!!! 
	if (tr[nw].l == tr[nw].r) {
		if (!b) {
			tr[nw].a = a; tr[nw].b = 0; tr[nw].sum = 0;
		} else {
			tr[nw].a = 0; tr[nw].b = b; tr[nw].sum = a * b;
		}
		return;
	} int mid = (tr[nw].l + tr[nw].r) >> 1;
	if (x <= mid) update(lson, x, a, b);
	else update(rson, x, a, b);
	pushup(nw);
}
pair<ll, ll> geta(int nw, int x, int y) {
	if (x <= tr[nw].l && tr[nw].r <= y) {
		return make_pair(tr[nw].a, tr[nw].b);
	} int mid = (tr[nw].l + tr[nw].r) >> 1;
	pair<ll, ll> l = {}, r = {};
	if (x <= mid) l = geta(lson, x, y);
	if (y > mid) r = geta(rson, x, y);
	if (l.second < r.first) return make_pair(l.first + r.first - l.second, r.second);
	else return make_pair(l.first, r.second + l.second - r.first);
}
ll ask(int nw, int x, ll k) { //return的值是nw里面实际上删除了多少个 
	//!!!!!!!!!!!不能直接左边取k个，因为可能右半部分会抵消
	if (!k) return 0;
	if (tr[nw].l == tr[nw].r) {
		if (!tr[nw].b) return 0;
		return min(tr[nw].b, k) * (tr[nw].sum / tr[nw].b);
	} int mid = (tr[nw].l + tr[nw].r) >> 1;
	if (x <= mid) return ask(lson, x, k);
	else {
		ll a = geta(rson, mid + 1, x).first;
		ll lcnt = max(tr[lson].b - a, 0ll);
		if (k <= lcnt) return ask(lson, x, k);
		else {
			return merge(lcnt, lson) + ask(rson, x, k - lcnt); 
		}
	}
}
int main() {
	n = read(), m = read();
	for (int i = 1; i <= m; i++) {
		int op = read(); ll l = read(), r = read(), x = read();
		if (op == 1) {
			ll y = read(); a[++tot] = Op(l, i, -1, x, y); 
			if (r != n) a[++tot] = Op(r + 1, i, -3, x, y); //先修改，再查询 
		} else if (op == 2) {
			a[++tot] = Op(l, i, -2, x, 0);
			if (r != n) a[++tot] = Op(r + 1, i, -3, x, 0);
		} else {
			++all; a[++tot] = Op(l, i, all, r, x);
		}
	}
	sort(a + 1, a + tot + 1, cmp); build(1, 1, m);
	for (int i = 1; i <= m; i++) {
		int x = a[i].x;
		while (i <= m && a[i].x == x) {
			if (a[i].t == -3) update(1, a[i].y, 0, 0);
			else if (a[i].t == -2) update(1, a[i].y, a[i].a, 0);
			else if (a[i].t == -1) update(1, a[i].y, a[i].b, a[i].a);  
			else {
				ans[a[i].t] = ask(1, a[i].y, a[i].b) - ask(1, a[i].y, a[i].a - 1); // 1~a[i].y次修改中， 前a[i].b个元素的和
			} i++;
		} i--;
	}
	for (int i = 1; i <= all; i++) printf("%lld\n", ans[i]);
    return 0;
}