// TLE 解法: 区間スケジューリングを愚直に N - K + 1 回する

#include <bits/stdc++.h>
using namespace std;
using P = pair<int, int>;

int main(){
    int N, K;
    cin >> N >> K;
    vector<int> L(N), R(N);
    vector<P> v(N);
    for(int i = 0; i < N; i++){
        cin >> L[i] >> R[i];
        v[i] = P(R[i], L[i]);
    }
    sort(v.begin(), v.end());
    const int inf = 1'000'000'100; // 1e9 + 100
    int ans = inf;
    for(int i = 0; i < N; i++){
        int cur = i, cnt = 1, next = i + 1;
        while(cnt != K && next < N){
            if(v[next].second >= v[cur].first){
                cur = next;
                cnt++;
            }
            next++;
        }
        if(cnt == K) ans = min(ans, v[cur].first - v[i].second);
    }
    if(ans != inf) cout << ans << endl;
    else cout << -1 << endl;
}