There are many freight trains departing from Kirnes planet every day. One day on that planet consists of h hours, and each hour consists of m minutes, where m is an even number. Currently, there are n freight trains, and they depart every day at the same time: i-th train departs at h_i hours and m_i minutes.

The government decided to add passenger trams as well: they plan to add a regular tram service with half-hour intervals. It means that the first tram of the day must depart at 0 hours and t minutes, where 0 ≤ t < {m \over 2}, the second tram departs m \over 2 minutes after the first one and so on. This schedule allows exactly two passenger trams per hour, which is a great improvement.

To allow passengers to board the tram safely, the tram must arrive k minutes before. During the time when passengers are boarding the tram, no freight train can depart from the planet. However, freight trains are allowed to depart at the very moment when the boarding starts, as well as at the moment when the passenger tram departs. Note that, if the first passenger tram departs at 0 hours and t minutes, where t < k, then the freight trains can not depart during the last k - t minutes of the day.

<image> A schematic picture of the correct way to run passenger trams. Here h=2 (therefore, the number of passenger trams is 2h=4), the number of freight trains is n=6. The passenger trams are marked in red (note that the spaces between them are the same). The freight trains are marked in blue. Time segments of length k before each passenger tram are highlighted in red. Note that there are no freight trains inside these segments.

Unfortunately, it might not be possible to satisfy the requirements of the government without canceling some of the freight trains. Please help the government find the optimal value of t to minimize the number of canceled freight trains in case all passenger trams depart according to schedule.

Input

The first line of input contains four integers n, h, m, k (1 ≤ n ≤ 100 000, 1 ≤ h ≤ 10^9, 2 ≤ m ≤ 10^9, m is even, 1 ≤ k ≤ {m \over 2}) — the number of freight trains per day, the number of hours and minutes on the planet, and the boarding time for each passenger tram.

n lines follow, each contains two integers h_i and m_i (0 ≤ h_i < h, 0 ≤ m_i < m) — the time when i-th freight train departs. It is guaranteed that no freight trains depart at the same time.

Output

The first line of output should contain two integers: the minimum number of trains that need to be canceled, and the optimal starting time t. Second line of output should contain freight trains that need to be canceled.

Examples

Input


2 24 60 15
16 0
17 15


Output


0 0



Input


2 24 60 16
16 0
17 15


Output


1 0
2 

Note

In the first test case of the example the first tram can depart at 0 hours and 0 minutes. Then the freight train at 16 hours and 0 minutes can depart at the same time as the passenger tram, and the freight train at 17 hours and 15 minutes can depart at the same time as the boarding starts for the upcoming passenger tram.

In the second test case of the example it is not possible to design the passenger tram schedule without cancelling any of the freight trains: if t ∈ [1, 15], then the freight train at 16 hours and 0 minutes is not able to depart (since boarding time is 16 minutes). If t = 0 or t ∈ [16, 29], then the freight train departing at 17 hours 15 minutes is not able to depart. However, if the second freight train is canceled, one can choose t = 0. Another possible option is to cancel the first train and choose t = 13.
#include<bits/stdc++.h>
#define SZ(x) ((int)x.size())
#define uni(x) sort(all(x)),x.resize(unique(all(x))-x.begin());
#define GETPOS(c,x) (lower_bound(all(c),x)-c.begin())
#define lown1(x,val) low(in(x),val)-x
#define lowm1(x,val) low(im(x),val)-x
#define low1(x,nums,val) low(x+1,x+nums+1,val)-x
#define mst(x,val) memset((x),val,sizeof((x)));
#define ls rt<<1
#define rs rt<<1|1
#define lson rt<<1,l,M
#define rson rt<<1|1,M+1,r
#define PI acos(-1)
#define MM int M=(l+r)>>1;
#define fu(i,r,t) for(int i=r;i<=t;i++)
#define fd(i,r,t) for(int i=r;i>=t;i--)
#define fh(i,be,e) for(int i=head[be];~i;i=e[i].next)
#define fa(i,V) for(auto i:V)
#define far(i,V) for(auto &i:V)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) (a*b)/__gcd(a,b)
#define cp(i,ans) printf("%.if",ans);
#define cpp(i,ans) cout<<setprecision(i)<<fixed<<ans<<endl;
#define ppb pop_back
#define ppf pop_front
#define pb push_back
#define pf push_front
#define pq priority_queue
#define lowbit(x) (x)&(-x)
#define all(V) V.begin(),V.end()
#define ms multiset
#define mod(x) ((x+mo_num)%mo_num)
#define vc vector
#define vct vector<int>
#define out(i) cout<<(i)<<endl;
#define fi first
#define se second
#define fun(i) fu(i,1,n)
#define fut(i) fu(i,1,t)
#define fum(i) fu(i,1,m)
#define ld long double
#define umap unordered_map
#define P pair<int,int>
#define SET set<int>
#define mp map<int,int>
#define mk make_tuple
#define eps 1e-6
//Remember cancel"#define endl '\n'" in interactive questions or use "<<flush"
#define endl '\n'
#define low lower_bound
#define upp upper_bound
#define yn(key) out(key?"YES":"NO")
//#define yn(key) out(key?"Yes":"No")
#define in(i) i+1,i+1+n
#define im(i) i+1,i+1+m
#define ik(i,k) i+1,i+1+k
#define bffs(i) __builtin_ffs(i)
#define bcount(i) __builtin_popcount(i)
#define bone(i) ((1<<i)-1)
#define db double
#define got(container,num) get<num-1>(container)
#define int long long
#define print(a,n) fun(i)cout<<a[i]<<(i!=n?' ':endl);
using namespace std;
//Remember to cancel the line below and declare INT=INT_MAX/2; when you want to change long to int
const int INF=LLONG_MAX/4,SINF=0x3f3f3f3f,Lim=1<<20,MINF=LLONG_MAX;
//const int INF=INT_MAX/4;
// use C:printf("%.16f", x);  ->  printf("%.10f", x); can accelerate the program
typedef unsigned long long ull; template<class T> void _R(T &x) { cin >> x; }void _R(int32_t &x) { scanf("%d", &x); }void _R(long long &x) { scanf("%lld", &x); }void _R(ull &x) { scanf("%llu", &x); }void _R(double &x) { scanf("%lf", &x); }void _R(char &x) { scanf(" %c", &x); }void _R(char *x) { scanf("%s", x); }void R() {}template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }template<class T> void _W(const T &x) { cout << x; }void _W(const int32_t &x) { printf("%d", x); }void _W(const long long &x) { printf("%lld", x); }void _W(const double &x) { printf("%.16f", x); }void _W(const char &x) { putchar(x); }void _W(const char *x) { printf("%s", x); }template<class T,class U> void _W(const pair<T,U> &x) {_W(x.F); putchar(' '); _W(x.S);}template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }void W() {}template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }
const int dx[]={0,0,-1,1},dy[]={-1,1,0,0};//up down left right
const int maxn=1e6+1e5;
const int mo_num=1e9+7;
//const int mo_num=998244353;
int n,m,t,a[maxn],b[maxn],ans;
//int f[maxn];//pos:下标;val:更新的值;up:数组容量上限,一般是n,getsum(n+1)是错误的;
umap<int,int>f;
const int up=1e9;
void update(int pos,int val){ for(int i=pos;i<=up;i+=lowbit(i)) f[i] += val; }

//求和A1~Apos
int getsum(int pos){ int num = 0;for(int i=pos;i;i-=lowbit(i))num += f[i];return num; }
//int state[maxn],num;
SET have;
void solve()
{
    int k;
    cin>>n>>a[1]>>m>>k;
    int half=m/2;
    fun(i)
    {
        int x;
        cin>>a[1]>>x;
        b[i]=x;
        update(1+x%half,1);
//        state[++num]=(x+1)%half;
        have.insert((x+1)%half);
    }
    if(k==1){
        out("0 0")
    }
    ans=INF;
    int T=0;
    fa(i,have)
    {
        t=i+k-1;
        int num=0;
        if(t-1>=half)
        {
            num=getsum(1+(t-1)%half)+getsum(1+half)-getsum(1+i-1);
        }else num=getsum(1+(t-1)%half)-(i>=1?getsum(1+i-1):0);
        if(num<ans)
        {
            ans=num;
            T=t;
        }
    }
    T%=half;
    cout<<(ans==INF?0:ans)<<" "<<T<<endl;
    fun(i)
    {
        int nn=b[i]%half;//T-k+1,T-1
        if(T-k+1<0)
        {
            if(nn<=T-1 or nn>=half+T-k+1)cout<<i<<" ";
        }else if(T-k+1<=nn && nn<=T-1)cout<<i<<" ";
    }
    return ;
}
main()
{
    IOS
    int T=1;
    //cin>>T;
    while(T--)solve();
    return 0;
}
