#include <bits/stdc++.h>
#define MAXS 3100000
#define MAX 100005
#define SEG3LIM (1000000007)
using ll = long long;
struct __attribute__((packed)) node {
int left;
int right;
ll sum;
int total;
};
node memoria[MAXS];
int cur=1;
int alocar(void){
return cur++;
}
int sz;
node nodes[MAX];
ll vals[MAX];
void inserir(node* x,ll val,int p,int la=0,int ra=SEG3LIM-1){
if(la==ra){
assert(p==la);
x->total++;
x->sum+=val;
return;
}
int m = (la+ra)/2;
if(m>=p){
auto old = x->left;
x->left=alocar();
memoria[x->left]=memoria[old];
inserir(&memoria[x->left],val,p,la,m);
}else {
auto old = x->right;
x->right=alocar();
memoria[x->right]=memoria[old];
inserir(&memoria[x->right],val,p,m+1,ra);
}
ll tot = 0,ss=0;
if(x->left)tot+=memoria[x->left].sum;
if(x->right)tot+=memoria[x->right].sum;
if(x->left)ss+=memoria[x->left].total;
if(x->right)ss+=memoria[x->right].total;
/// std::cout<<"Node "<<tot<<" "<<ss<<"\n";
x->sum=tot;
x->total=ss;
}
long long ans=0,restante=0;
void walk(node* k,node* u,int la=0,int ra=SEG3LIM-1){
if(!restante||!(k->sum))return;
long long subsum=0,subtotal=0;
subsum+=u->sum;
subtotal+=u->total;
if(k->total-subtotal<=restante){
ans+=k->sum-subsum;
restante-=k->total-subtotal;
/// std::cout<<"Add "<<k->sum-subsum<<" "<<la<<" "<<ra<<" "<<restante<<"\n";
return;
}
int m=(la+ra)/2;
int tem = 0;
if(k->right){
tem=memoria[k->right].total-memoria[u->right].total;
}
if(la==ra){
ll pega = std::min(restante,k->total-subtotal);
restante-=pega;
ans+=((long long)la)*pega;
/// std::cout<<"Especial "<<pega<<"\n";
return;
}
if(la==ra)return;
walk(&memoria[k->right],&memoria[u->right],m+1,ra);
walk(&memoria[k->left],&memoria[u->left],la,m);
}
node locais[MAX];
typedef std::pair<ll,ll> pll;
///Eu sei como resolver para um M infinito, mas como faco para um M FINITO???
///Ideia do M infinito:
///O custo do absoluto eh basicamente MAX*2 - MIN*2
///O resto N importa muito, eu tenho que pegar a soma dos M-2 maiores elementos entre L e R (alem da soma do A de L e R)
///Relaxacao lagrangiana talvez?
///Dividir para conquistar?
int N,M;
const ll inf = (1LL<<60LL);
std::vector<pll> vec;
long long pode=0;
long long resp=-inf;
long long consulta(int l,int r,int p){///Naive
{
std::vector<long long> v;
for(int i=l;i<=r;++i){
v.push_back(vec[i].second);
}
std::sort(v.begin(),v.end(),std::greater<long long>());
if(v.size()<p)return -inf;
long long ans=0;
for(int j=0;j!=std::min((int)v.size(),p);++j){
ans+=v[j];
}
return ans;
}
if(r<l)return -inf;
if(r-l+1<p)return -inf;
ans=0;restante=p;
auto beta = &locais[r],alpha=&memoria[0];
if(l){
alpha=&locais[l-1];
}
walk(beta,alpha);
return ans;
}
int count=0;
void dp(int l,int r,int la,int ra){
if(r<l)return;
int m = (l+r)/2;
ll ans = -inf,split=la;
++count;
for(int i=la;i<=std::min(ra,m-1);++i){
ll que = consulta(i+1,m-1,M-2);
ll bonus = que-(2LL*vec[m].first)+vec[m].second+vec[i].second+(2LL*vec[i].first);
///std::cout<<bonus<<" "<<que<<" "<<i+1<<" "<<m-1<<"!\n";
if(bonus>=ans){
ans=bonus;
split=i;
}
}
resp=std::max(resp,ans);
dp(l,m-1,la,split);
dp(m+1,r,split,ra);
}
int main()
{
std::cin>>N>>M;
for(int i=0;i!=N;++i){
ll a,b;
std::cin>>a>>b;
vec.push_back({b,a});
}
std::sort(vec.begin(),vec.end());
for(int i=0;i!=N;++i){
/// std::cout<<vec[i].first<<" "<<vec[i].second<<"\n";
if(i)locais[i]=locais[i-1];
///Comprimir cordenada!!!!!!!! N sei se precisa, provavelmente n, a arvore eh esparsa
inserir(&locais[i],vec[i].second,vec[i].second);
}
dp(0,N-1,0,N-1);
assert(count==N);
std::cout<<resp<<"\n";
}
