#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;

// デバッグ表示
#define dump(x) cout << #x << ":" << (x) << endl;

// 型定義
typedef long long ll;
typedef pair<ll, ll> P;

// forループ
#define REP(i,n) for(ll i=0; i<(ll)(n); ++i)

// 定数宣言
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

// modint
using mint = modint1000000007;
// using mint = modint998244353;

// グラフ表現
using Graph = vector<vector<int>>;

// グラフの辺表現
using Edge = map<pair<int,int>,int>;

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

// コンビネーションを計算する関数
ll pow(ll N, ll k) {
    ll res = 1;
    for (ll i = 0; i < k; ++i) res *= N;
    return res;
}

// 最大公約数
ll gcd(ll a,ll b){
   if (a%b == 0) return(b);
   else return(gcd(b, a%b));
}

// 最小公倍数
ll lcm(ll a, ll b){
    return a/gcd(a, b) * b;
}

int main()
{
    cout << fixed << setprecision(15);
    ll N, D;
    cin >> N >> D;

    vector<ll> p(N);
    vector<ll> q(N);

    REP(i, N){
        cin >> p[i] >> q[i];
    }

    ll ans = -LINF;
    REP(i, N){
        REP(j, N){
            if(i == j) continue;
            ll now = 0;
            ll temp_min = LINF;

            now -= p[i];
            temp_min = min(temp_min, now);
            now += q[i];

            now -= p[j];
            temp_min = min(temp_min, now);
            now += q[j];

            if(D == 2){
                ans = max(ans, temp_min);
                continue;
            }

            now -= p[i];
            temp_min = min(temp_min, now);
            if(D == 3){
                ans = max(ans, temp_min);
                continue;
            }
            now += q[i];

            now -= p[j];
            temp_min = min(temp_min, now);
            if(D == 3){
                ans = max(ans, temp_min);
                continue;
            }
            now += q[j];

            ll k = (D-4);
            ll t = -p[i] + q[i] -p[j] + q[j];
                
            now += k * t;

            now -= p[i];
            temp_min = min(temp_min, now);
            if(D%2 == 1){
                ans = max(ans, temp_min);
                continue;
            }
            now += q[i];

            now -= p[j];
            temp_min = min(temp_min, now);
            if(D%2 == 0){
                ans = max(ans, temp_min);
                continue;
            }
            now += q[j];

        }
    }

    cout << ans << endl;
    return 0;
}