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;
using Long = __int128_t;

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

    int n, q;
    long long l, r;
    cin >> n >> q >> l >> r;
    vector<int> as(n);
    for (int i = 0; i < n; i++) cin >> as[i];

    vector<long long> qs(q), xs(q), ss(q), ts(q);
    for (int i = 0; i < q; i++) {
        cin >> qs[i] >> xs[i] >> ss[i] >> ts[i];
    }

    Long lmima = -((Long) 1 << 64), lijou = (Long) 1 << 64;
    while (lmima < lijou - 1) {
        Long m = (lmima + lijou) / 2;
        Long v = m;
        for (int i = 0; i < q; i++) {
            if (qs[i] == 1) {
                if (v >= xs[i]) {
                    v = ts[i] * (v + ss[i]);
                }
            } else {
                if (v <= xs[i]) {
                    v = (v - ss[i]) / ts[i];
                }
            }
        }
        if (v >= l) {
            lijou = m;
        } else {
            lmima = m;
        }
    }

    Long rika = -((Long) 1 << 64), rue = (Long) 1 << 64;
    while (rika < rue - 1) {
        Long m = (rika + rue) / 2;
        Long v = m;
        for (int i = 0; i < q; i++) {
            if (qs[i] == 1) {
                if (v >= xs[i]) {
                    v = ts[i] * (v + ss[i]);
                }
            } else {
                if (v <= xs[i]) {
                    v = (v - ss[i]) / ts[i];
                }
            }
        }
        if (v <= r) {
            rika = m;
        } else {
            rue = m;
        }
    }

    int cnt = 0;
    for (int i = 0; i < n; i++) {
        if (as[i] >= lijou && as[i] <= rika) cnt++;
    }
    cout << cnt << endl;
}