【題目連結】
　　http://uoj.ac/problem/386
【題解】
　　考慮先把序列按S$S$排序後，將V$V$從小到大從中刪去，刪去時用包含它的區間更新答案。
　　為什麼答案一定是一段區間：由於數列中剩下的數V$V$沒有比當前數更小的。用反證法，如果不是一段區間，那麼把當前點替換為區間隔開的點答案不會變劣。
　　時間複雜度O(NM2)$O(N{M}^2)$
【程式碼】

/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :   [UNR3Day1T1]
    Points :
- - - - - - - - - - - - - - - */
 

# include <bits/stdc++.h>
# define    ll      long long
# define    N       200010
using namespace std;
const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
const ll  infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'
 
){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
struct Node{ll s, v;}p[N];
bool cmp(Node x, Node y){return x.s < y.s;}
priority_queue <ll, vector<ll>, greater<ll> > hp;
int
 
 n, m, S, V;
ll ans;
void chkans(ll sum, ll det){
    ll num1, num2;
    if (V == 1) num1 = sum; else num1 = sum * sum;
    if (S == 1) num2 = det; else num2 = det * det;
    ans = max(ans, num1 - num2);
}
int main(){
    n = read(), m = read(), S = read(), V = read();
    for (int i = 1; i <= n; i++)
        p[i].s = read(), p[i].v = read();
    sort(p + 1, p + n + 1, cmp);
    ans = -INFll;
    for (int i = 1; i <= n; i++){
        while (hp.size() > 0) hp.pop();
        ll now = 0;
        for (int j = i; j <= n; j++){
            if (hp.size() == m){
            if (hp.top() < p[j].v){
                now = now - hp.top() + p[j].v;
                hp.pop();
                hp.push(p[j].v);
            }
            }
            else {
                now = now + p[j].v;
                hp.push(p[j].v);
            }
            chkans(now, p[j].s - p[i].s);
        }
    }
    printf("%lld\n", ans);
    return 0;
}