#include using namespace std; typedef long long ll; typedef long double ld; typedef vector vi; typedef vector vd; typedef vector vll; typedef vector vs; typedef vector vb; typedef pair pii; struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star; #define all(c) (c).begin(), (c).end() #define rall(c) (c).rbegin(), (c).rend() #define vv(type, c, m, n, i) vector> c(m, vector(n, i)) #define mat(type, c, m, n) vector> c(m, vector(n));for(auto& r:c)for(auto& i:r)cin>>i; #define rep(i,a,n) for(int i=(a), i##_len=(n); ii##_len; --i) #define each(x,y) for(auto &(x):(y)) #define var(type, ...)type __VA_ARGS__;Scan(__VA_ARGS__); template void Scan(T& t) { cin >> t; } templatevoid Scan(First& first, Rest&...rest) { cin >> first; Scan(rest...); } #define vec(type, c, n) vector c(n);for(auto& i:c) cin>>i; #define len(x) ((int)(x).size()) #define mp make_pair #define eb emplace_back #define em emplace #define pb pop_back #define fi first #define se second #define get(a, i) get(a) #define bit_count __builtin_popcount #define mt make_tuple inline string zero(int a,string s,char c='0'){return string(a-len(s),c)+s;} inline string zero(int a,int t,char c='0'){return zero(a, to_string(t), c);} inline string zero(int a,ll t,char c='0'){return zero(a, to_string(t), c);} #define mini min_element #define maxi max_element #define sum accumulate #define chmax(a,b) if(ab)a=b template inline void print(T t){cout << t << '\n';} template inline void print(H h, T... t){cout << h << " ";print(t...);} int n, ans = 1000; vector A; map m; void f(int num, int l, int max_A) { if (max_A >= ans) { return; } if (l == 0) { ans = max_A; return; } each(i, m[num]) { if ((1 << i) & l) { chmax(max_A, A[num].se + A[i].fi); f(i, l - (1 << i), max_A); } } } void solve() { cin >> n; int a, b; rep(i, 0, n) { cin >> a >> b; A.eb(mp(a, b - a)); } sort(all(A)); int ans1 = 0, ans2 = 0; rep(i, 1, n) { chmax(ans1, A[i - 1].se + A[i].fi); chmax(ans2, A[i - 1].fi + A[i].se); } ans = min({ans, ans1, ans2}); rep(i, 0, n) { rep(j, 0, n) { int x = A[i].se + A[j].fi; if (x <= ans) { m[i].eb(j); } } } rep(i, 0, n) { f(i, ((1 << n) - 1) - (1 << i), 0); } } int main() { solve(); print(ans); return 0; }