/*******************************************
Task: JOI19_Cake3
Link: https://oj.uz/problem/view/JOI19_cake3
*******************************************/
#include <bits/stdc++.h>
#define maxn 200005
#define fi first
#define se second
using namespace std;
using ii = pair<int, int>;
using li = pair<int64_t, int>;
int n, k;
ii a[maxn];
template<class T> inline constexpr void cmax(T &x, const T &y) {if (x < y) x = y;}
int64_t sum[22*maxn];
int b[maxn], n1 = 0, root[maxn];
int it[22*maxn], L[22*maxn], R[22*maxn], nt = 0;
int build(int lo = 1, int hi = n1) {
if (lo == hi) return ++nt;
int cur = ++nt, mid = (lo + hi) >> 1;
L[cur] = build(lo, mid);
R[cur] = build(mid+1, hi);
return cur;
}
int upd(int u, int oldver, int lo = 1, int hi = n1) {
if (lo == hi) {
++nt;
it[nt] = it[oldver] + 1;
sum[nt] = sum[oldver] + b[lo];
return nt;
}
int cur = ++nt, mid = (lo + hi) >> 1;
if (u <= mid) {
L[cur] = upd(u, L[oldver], lo, mid);
R[cur] = R[oldver];
} else {
L[cur] = L[oldver];
R[cur] = upd(u, R[oldver], mid+1, hi);
}
it[cur] = it[L[cur]] + it[R[cur]];
sum[cur] = sum[L[cur]] + sum[R[cur]];
return cur;
}
int64_t kth(int k, int v1, int v2, int lo = 1, int hi = n1) {
if (lo == hi)
return 1LL * k * b[lo];
int mid = (lo + hi) >> 1, phai = it[R[v2]]-it[R[v1]];
if (k > phai) return kth(k-phai, L[v1], L[v2], lo, mid) + sum[R[v2]] - sum[R[v1]];
return kth(k, R[v1], R[v2], mid+1, hi);
}
int64_t kq[maxn];
void solve(int l = 1, int r = n, int optl = 1, int optr = n) {
int m = (l + r) / 2;
li Best = {LLONG_MIN, 0};
for (int i = max(m+k-1, optl); i <= optr; i++) cmax(Best, li{kth(k, root[m-1], root[i]) - 2 * (a[i].se-a[m].se), i});
if (!Best.se) return;
kq[m] = Best.fi; int opt = Best.se;
if (l < m) solve(l, m-1, optl, opt);
if (r > m) solve(m+1, r, opt, optr);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cin >> n >> k;
for (int i = 1; i <= n; i++) cin >> a[i].fi >> a[i].se;
sort(a + 1, a + n + 1, [&](ii x, ii y) {return x.se < y.se;});
for (int i = 1; i <= n; i++) b[++n1] = a[i].fi;
sort(b + 1, b + n1 + 1);
n1 = unique(b + 1, b + n1 + 1) - b - 1;
root[0] = build();
for (int i = 1; i <= n; i++) {
int p = lower_bound(b + 1, b + n1 + 1, a[i].fi) - b;
root[i] = upd(p, root[i-1]);
}
for (int i = 1; i <= n; i++) kq[i] = LLONG_MIN;
solve();
cout << *max_element(kq + 1, kq + n + 1);
}
