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 long long ll;

struct query{
    long long q,x,s,t;
};

ll N, Q, L, R;
vector<long long> A;
vector<query> queries;

long long solve();
long long calc(long long now);

int main(){
    cin >> N >> Q >> L >> R;
    A.resize(N);
    for(int i = 0;i < N;++i)cin >> A[i];
    queries.resize(Q);
    for(int i = 0;i < Q;++i)
        cin >> queries[i].q >> queries[i].x >> queries[i].s >> queries[i].t;
    cout << solve() << endl;
    return 0;
}

long long solve(){
    sort(A.begin(),A.end());
    long long l = 0,r = N - 1,ansl = 0,ansr = N-1;
    if(L <= calc(A[0]))r = 0;
    else if(calc(A[N-1]) < L)return 0;
    else
        while(r - l > 1){
            long long now = (l + r)/2,nowans = 0;
            nowans = calc(A[now]);
            if(L <= nowans)r = now;
            else l = now;
        }
    ansl = r;
    l = 0;
    r = N-1;
    if(R >= calc(A[N-1]))l = N-1;
    else if(calc(A[0]) > R)return 0;
    else
        while(r - l > 1){
            long long now = (l + r)/2,nowans = 0;
            nowans = calc(A[now]);
            if(R >= nowans)l = now;
            else r = now;
        }
        ansr = l;
    return ansr - ansl + 1;
}

long long calc(long long now){
    for(int i = 0;i < Q;++i){
        long long x = queries[i].x,s = queries[i].s,t = queries[i].t;
        if(queries[i].q == 1){
            if(x <= now)now = (now + s) * t;
        }
        else{
            if(x >= now)now = (now - s) / t;
        }
    }
    return now;
}