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;
#define rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define double long double


#define ll long long
using box = vector<ll>;


ll check(ll v1, box& q, box& x, box& s, box& t)
{
    int Q = q.size();
    ll v = v1;
    rep(i, Q)
    {
        if(q[i] == 1)
        {
            if(v >= x[i])
            {
                v = t[i] * (v + s[i]);
            }
        }
        else
        {
            if(v <= x[i])
            {
                v = (v - s[i]) / t[i]; 
            }
        }
        //cout << v << endl;
    }
    return v;
}
void solve()
{
    ll N, Q, L, R;
    cin >> N >> Q >> L >> R;
    vector<ll> A(N);
    rep(i, N) cin >> A[i];
    sort(A.begin(), A.end());
    vector<ll> q(Q);
    vector<ll> x(Q);
    vector<ll> s(Q);
    vector<ll> t(Q);
    rep(i, Q)
    {
        cin >> q[i] >> x[i] >> s[i] >> t[i];
    }
    //L以上となる最初の頂点
    int ok = N;
    int ng = -1;
    while(abs(ok - ng) > 1)
    {
        int mid = (ok + ng) / 2;
        if(check(A[mid], q, x, s, t) >= L)
        {
            ok = mid;
        }
        else{
            ng = mid;
        }
    }
    //cout << ok << " " << ng << endl;

    //R以下となる最大の頂点
    int rok = -1;
    int rng = N;
    while(abs(rok - rng) > 1)
    {
        int mid = (rok + rng) / 2;
        if(check(A[mid], q, x, s, t) <= R)
        {
            rok = mid;
        }
        else{
            rng = mid;
        }
    }
    //cout << rok << " " << rng << endl;

    cout << (max(rok - ok + 1, 0)) << endl;

    // rep(i, N){
    //     cout << A[i] << " " << (check(A[i], q, x, s, t)) << endl;
    // }

}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}