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>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define SIZE 200005

struct Query{

	ll q,x,s,t;
};

ll N,Q,minimum,maximum;
ll A[SIZE];
Query query[SIZE];


int main(){

	scanf("%lld %lld %lld %lld",&N,&Q,&minimum,&maximum);

	for(int i = 0; i < N; i++){

		scanf("%lld",&A[i]);
	}

	sort(A,A+N);

	for(int i = 0; i < Q; i++){

		scanf("%lld %lld %lld %lld",&query[i].q,&query[i].x,&query[i].s,&query[i].t);
	}

	int left = N;
	int L = 0,R = N-1,mid = (L+R)/2;

	while(L <=R){

		ll tmp = A[mid];

		for(int i = 0; i < Q; i++){

			if(query[i].q == 1){ //AddMul

				if(tmp < query[i].x)continue;

				tmp += query[i].s;
				tmp *= query[i].t;

			}else{ //SubDiv

				if(tmp > query[i].x)continue;

				tmp -= query[i].s;
				tmp /= query[i].t;
			}
		}

		if(tmp >= minimum){

			left = mid;
			R = mid-1;

		}else{

			L = mid+1;
		}
		mid = (L+R)/2;
	}

	int right = -1;
	L = 0,R = N-1,mid = (L+R)/2;

	while(L <= R){

		ll tmp = A[mid];

		for(int i = 0; i < Q; i++){

			if(query[i].q == 1){ //AddMul

				if(tmp < query[i].x)continue;

				tmp += query[i].s;
				tmp *= query[i].t;

			}else{ //SubDiv

				if(tmp > query[i].x)continue;

				tmp -= query[i].s;
				tmp /= query[i].t;
			}
		}

		if(tmp <= maximum){

			right = mid;
			L = mid+1;

		}else{

			R = mid-1;
		}
		mid = (L+R)/2;
	}

	printf("%d\n",max(0,right-left+1));

	return 0;
}