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 <iostream>
#include <vector>
#include <algorithm>
#define rep(i, a) for (int i = 0; i < (a); i++)
#define int long long
using namespace std;

struct Q {
	int q, x, s, t;
};

int q, r, l;
vector<int> a;
vector<Q> que;

int cal(int mid){
	int cur = a[mid];
	rep(i,q){
		if(que[i].q == 1){
			if(cur >= que[i].x){
				cur = (cur + que[i].s) * que[i].t;
			}
		}else{
			if(cur <= que[i].x){
				cur = (cur - que[i].s) / que[i].t;
			}
		}
	}
	return cur;
}

bool isOk_max(int mid) {
	int cur = a[mid];
	rep(i,q){
		if(que[i].q == 1){
			if(cur >= que[i].x){
				cur = (cur + que[i].s) * que[i].t;
			}
		}else{
			if(cur <= que[i].x){
				cur = (cur - que[i].s) / que[i].t;
			}
		}
	}
	return cur < r;
}
bool isOk_min(int mid) {
	int cur = a[mid];
	rep(i,q){
		if(que[i].q == 1){
			if(cur >= que[i].x){
				cur = (cur + que[i].s) * que[i].t;
			}
		}else{
			if(cur <= que[i].x){
				cur = (cur - que[i].s) / que[i].t;
			}
		}
	}
	return cur > l;
}

signed main() {
	int n;
	cin >> n >> q >> l >> r;
	r++;
	l--;

	a = vector<int>(n);
	rep(i, n) cin >> a[i];
	sort(a.begin(), a.end());

	que = vector<Q>(q);
	rep(i, q) {
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		que[i] = {a, b, c, d};
	}

	// 上限を探す
	int ok = -1, ng = n;
	rep(i, 100) {
		int mid = (ok + ng) / 2;
		if (isOk_max(mid))
			ok = mid;
		else
			ng = mid;
	}

	int maxi = ok;

	ok = n, ng = -1;
	rep(i,100){
		int mid = (ok + ng) / 2;
		if (isOk_min(mid))
			ok = mid;
		else
			ng = mid;
	}

	//rep(i,n){ cout << cal(i) << ' '; } cout << endl;

	cout << maxi - ok + 1 << endl;
}