Problem: B: AddMulSubDiv

Problem Statement

You have an array A of N integers. A_i denotes the i-th element of A.

You have to process one of the following queries Q times:

* Query 1: The query consists of non-negative integer x, and two positive integers s, t. For all the elements greater than or equal to x in A, you have to add s to the elements, and then multiply them by t. That is, an element with v (v \geq x) becomes t(v + s) after this query.


* Query 2: The query consists of non-negative integer x, and two positive integers s, t. For all the elements less than or equal to x in A, you have to subtract s from the elements, and then divide them by t. If the result is not an integer, you truncate it towards zero. That is, an element with v (v \leq x) becomes $\mathrm{trunc}$ ( \frac{v - s}{t} ) after this query, where $\mathrm{trunc}$ ( y ) is the integer obtained by truncating y towards zero.


* "Truncating towards zero" means converting a decimal number to an integer so that if x = 0.0 then 0 , otherwise an integer whose absolute value is the maximum integer no more than |x| and whose sign is same as x. For example, truncating 3.5 towards zero is 3, and truncating -2.8 towards zero is -2.



After applying Q queries one by one, how many elements in A are no less than L and no more than R?

Input


N Q L R
A_1 A_2 ... A_N
q_1 x_1 s_1 t_1
:
q_Q x_Q s_Q t_Q


* The first line contains four integers N, Q, L, and R. N is the number of elements in an integer array A, Q is the number of queries, and L and R specify the range of integers you want to count within after the queries.
* The second line consists of N integers, the i-th of which is the i-th element of A.
* The following Q lines represent information of queries. The j-th line of them corresponds to the j-th query and consists of four integers q_j, x_j, s_j, and t_j. Here, q_j = 1 stands for the j-th query is Query 1, and q_j = 2 stands for the j-th query is Query 2. x_j, s_j, and t_j are parameters used for the j-th query.



Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq Q \leq 2 \times 10^5
* -2^{63} < L \leq R < 2^{63}
* 0 \leq |A_i| \leq 10^9
* q_j \in \\{ 1, 2 \\}
* 0 \leq x_j < 2^{63}
* 1 \leq s_j, t_j \leq 10^9
* Inputs consist only of integers.
* The absolute value of any element in the array dosen't exceed 2^{63} at any point during query processing.



Output

Output the number of elements in A that are no less than L and no more than R after processing all given queries in one line.

Sample Input 1


3 3 3 10
1 -2 3
1 2 2 3
2 20 1 3
2 1 20 5


Output for Sample Input 1


1





Example

Input

3 3 3 10
1 -2 3
1 2 2 3
2 20 1 3
2 1 20 5


Output

1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef __int128 ll;
typedef vector<ll> vl;

ll n,q,l,r;
vl a;
vl query,x,s,t;
ll ans_l,ans_r,ans;

bool judge(ll a,ll thresh){
	for(ll i = 0;i < q;i++){
		if(query[i] == 1){
			if(a >= x[i]){
				a = t[i]*(a+s[i]);
			}
		}else if(query[i] == 2){
			if(a <= x[i]){
				a = (a-s[i])/t[i];
			}
		}
	}
	if(a >= thresh) return true;
	else return false;
}

int main(){
	// cerr << ((ll)1 << 63)-1 << endl;
	// cerr << (((ll)1 << 63)-1)*-1 << endl;
	long long N,Q,L,R;
	cin >> N >> Q >> L >> R;
	n = N;
	q = Q;
	l = L;
	r = R;
	a = vl(n);
	for(ll i = 0;i < n;i++){
		long long tmp;
		cin >> tmp;
		a[i] = tmp;
	}
	// for(ll i = 0;i < n;i++) cin >> a[i];
	query = x = s = t = vl(q);
	for(ll i = 0;i < q;i++){
		long long hoge,fuga,zaki,o;
		cin >> hoge >> fuga >> zaki >> o;
		query[i] = hoge;
		x[i] = fuga;
		s[i] = zaki;
		t[i] = o;
	}
	// for(ll i = 0;i < q;i++) cin >> query[i] >> x[i] >> s[i] >> t[i];
	ll top = ((ll)1 << 63)-1;
	ll bottom = top*-1;
	while(top-bottom > 1){
		ll middle = (top+bottom)/2;
		if(judge(middle,l)) top = middle;
		else bottom = middle;
	}
	ans_l = top;
	top = ((ll)1 << 63)-1;
	bottom = top*-1;
	while(top-bottom > 1){
		ll middle = (top+bottom)/2;
		if(judge(middle,r+1)) top = middle;
		else bottom = middle;
	}
	ans_r = top;
	// cerr << (long long)ans_l << " " << (long long)ans_r << endl;
	sort(a.begin(),a.end());
	cout << lower_bound(a.begin(),a.end(),ans_r)-lower_bound(a.begin(),a.end(),ans_l) << endl;
}