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<vector>
#include<iostream>
#include<string>
#include<algorithm>
#include<math.h>
#include<map>
#include<functional>
#include<queue>
#include<stack>
#include<string.h>
#include<list>
#include<limits>
#include<bitset>
#include<ctype.h>
#include<set>
#define all(vec) vec.begin(),vec.end()
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
int q;
vector<ll> qs,x,s,t;
ll f(ll b){
    for(int i=0;i<q;i++){
            if(qs[i]==1){
                if(b>=x[i]){
                    b+=s[i];
                    b*=t[i];
                }
            }else{
                if(b<=x[i]){
                    b-=s[i];
                    b/=t[i];
                }
            }
        }
    return b;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,l,r;cin>>n>>q>>l>>r;
    vector<ll> a(n);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(all(a));
    qs.resize(q);
    x.resize(q);
    s.resize(q);
    t.resize(q);
    for(int i=0;i<q;i++){
        cin>>qs[i]>>x[i]>>s[i]>>t[i];
    }
    int ok=0,ng=n;
    while(ng-ok>1){
        int mid=(ok+ng)/2;
        if(f(a[mid])<l){
            ok=mid;
        }else{
            ng=mid;
        }
    }
    int lb=ok;
    if(f(a[lb])>=l){
        --lb;
    }
    ok=0,ng=n;
    while(ng-ok>1){
        int mid=(ok+ng)/2;
        if(f(a[mid])<=r){
            ok=mid;
        }else{
            ng=mid;
        }
    }
    int rb=ok;
    if(f(a[rb])>r){
        --rb;
    }
    cout<<(rb-lb)<<endl;
}