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) for(ll (i) = (0);(i) < (n);++i)
#define ALL(v) v.begin(),v.end()
#define FI first
#define SE second

typedef long long ll;
typedef pair<ll, ll> P;
ll n, q, l, r;

ll seica(ll num, vector<pair<P, P>> &que) {
    REP(i, q){
        ll x = que[i].FI.SE;
        ll s = que[i].SE.FI;
        ll t = que[i].SE.SE;
        if(que[i].FI.FI == 1){
            if(num >= x){
                num = t * (num + s);
            }
        }
        else {
            if(num <= x){
                num = (num - s) / t;
            }
        }
    }
    return num;
}

bool checkMIN(ll num, vector<pair<P, P>> &que) {
    num = seica(num, que);
    return num < l;
}

bool checkMAX(ll num, vector<pair<P, P>> &que) {
    num = seica(num, que);
    return num <= r;
}

int main() {
    cin >> n >> q >> l >> r;
    vector<ll> v(n);REP(i, n)cin >> v[i];
    vector<pair<P, P>> que(q);
    REP(i, q){
        cin >> que[i].FI.FI >> que[i].FI.SE >> que[i].SE.FI >> que[i].SE.SE;
    }

    sort(ALL(v));
    ll L = -1, R = n;
    while(R - L > 1){
        ll mid = (L + R) / 2;
        //cout << "mim " << L << " " << R << " " << mid << endl;
        if(checkMIN(v[mid], que))L = mid;
        else R = mid;
    }
    ll ans_l = L;
    L = -1;R = n;
    while(R - L > 1){
        ll mid = (L + R) / 2;
        //cout << "mim " << L << " " << R << " " << mid << endl;
        if(checkMAX(v[mid], que))L = mid;
        else R = mid;
    }
    ll ans_r = L;
    //cout << "ans " << ans_l << " " << ans_r << endl;

    cout << max(0LL, ans_r - ans_l) << endl;
    return 0;
}